Permutations and Combinations DPP(bansal)

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MATHEMATICSDaily Practice Problems

Target IIT JEE 2010

ELEMENTARY PROBLEMS ON PERMUTATION & COMBINATIONNOTE: USE FUNDAMENTAL PRINCIPLE OF COUNTING & ENJOY DOING THE FOLLOWING.Q.11/1 In how many ways can clean & clouded (overcast) days occur in a week assuming that an entire day is

either clean or clouded. [ Ans. 27 = 128 ]Q.22/1 Four visitors A, B, C & D arrive at a town which has 5 hotels. In how many ways can they disperse

themselves among 5 hotels, if 4 hotels are used to accommodate them. [ Ans. 5 . 4 . 3 . 2 = 120]Q.33/1 If the letters of the word VARUN are written in all possible ways and then are arranged as in a

dictionary, then the rank of the word VARUN is :(A) 98 (B) 99 (C*) 100 (D) 101

Q.44/1 How many natural numbers are their from 1 to 1000 which have none of their digits repeated.[Hint: S D = 9 ; DD = 9 9 = 81 ; TD = 9 9 8 = 648] [ Ans. 738]Q.5 3 different railway passes are allotted to 5 students. The number of ways this can be done is :

(A*) 60 (B) 20 (C) 15 (D) 10Q.66/1 There are 6 roads between A & B and 4 roads between B & C.(i) In how many ways can one drive from A to C by way of B ?(ii) In how many ways can one drive from A to C and back to A, passing through B on both trips ?(iii) In how many ways can one drive the circular trip described in (ii) without using the same road more

than once. [ Ans. (i) 24 ; (ii) 576 ; (iii) 360 ]

Q.77/1 (i) How many car number plates can be made if each plate contains 2 different letters of Englishalphabet, followed by 3 different digits.

(ii) Solve the problem, if the first digit cannot be 0. (Do not simplify)[ Ans. (i) 26 . 25 . 10 . 9 . 8 = 468000 ; (ii) 26 . 25 . 9 . 9 . 8 = 421200 ]

Q.88/1 (i) Find the number of four letter word that can be formed from the letters of the word HISTORY.(each letter to be used at most once) [ Ans. 7 . 6 . 5 . 4 = 42 x 20 = 840 ]

(ii) How many of them contain only consonants? [ Ans. 5 . 4 . 3 . 2 = 120 ](iii) How many of them begin & end in a consonant? [ Ans. 5 . 4 . 4 . 4 = 400 ](iv) How many of them begin with a vowel? [ Ans. 240 ](v) How many contain the letters Y? [ Ans. 480 ](vi) How many begin with T & end in a vowel? [ Ans. 40 ](vii) How many begin with T & also contain S? [ Ans. 60 ](viii) How many contain both vowels? [ Ans. 240 ]

Q.99/1 If repetitions are not permitted(i) How many 3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 & 9 ? [ Ans. 120 ](ii) How many of these are less than 400 ? [ Ans. 40 ](iii) How many are even ? [ Ans. 40 ](iv) How many are odd ? [ Ans. 80 ](v) How many are multiples of 5 ? [ Ans. 20 ]Q.10 How many two digit numbers are there in which the tens digit and the units digit are different and odd?

[ Ans. 5 . 4 = 20 ]Q.1111/1 Every telephone number consists of 7 digits. How many telephone numbers are there which do not

include any other digits but 2 , 3 , 5 & 7 ? [ Ans. 47 ]

CLASS : XIII (VXYZ) Special DPP on Permutation and Combination DPP. NO.-1

Dpp-1 to 7

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Q.1212/1 (a) In how many ways can four passengers be accommodate in three railway carriages, if eachcarriage can accommodate any number of passengers.

(b) In how many ways four persons can be accommodated in 3 different chairs if each person canoccupy only one chair. [ Ans. (a) 34 ; (b) 4.3.2 = 24 ]

Q.13 How many odd numbers of five distinct digits can be formed with the digits 0,1,2,3,4 ?[ Hint: 3 3 2 1 2 = 36 Ans ]

Q.1414/1 Number of natural numbers between 100 and 1000 such that at least one of their digits is 7, is(A) 225 (B) 243 (C*) 252 (D) none

[Hint: 9 10 10 = 900 [ 13th Test (5-12-2004)]

Total no. of numbers without 7 = 81 8 = 648

required number = 900 648 = 252 ]Q.1515/1 How many four digit numbers are there which are divisible by 2. [Ans. = 4500]Q.16 The 120 permutations of MAHES are arranged in dictionary order, as if each were an ordinary five-letter

word. The last letter of the 86th word in the list is(A) A (B) H (C) S (D*) E [12th (14-5-2006)]

[Sol. The first 24 = 4! words begins with A, the next 24 begin with E and the next 24 begin with H. So the 86thbegins with M and it is the 86 72 = 14th such word. The first 6 words that begin with M begin with MAand the next 6 begin with ME. So the desired word begins with MH and it is the second such word. Thefirst word that begins with MH is MHAES, the second is MHASE. Thus E is the letter we seek. ]

Q.1717/1 Find the number of 7 lettered palindromes which can be formed using the letters from the Englishalphabets. [Ans. 264]

[Hint: A palindrome is a word or a phase that is the same whether you read is forward or backword.e.g. REFER]

Q.1818/1 Number of ways in which 7 different colours in a rainbow can be arranged if green is always in themiddle. [ Ans 720 ]

Q.1919/1 Two cards are drawn one at a time & without replacement from a pack of 52 cards. Determine thenumber of ways in which the two cards can be drawn in a definite order. [Ans. 52 51= 2652]

Q.20 Find the number of ways in which the letters of the word "MIRACLE" can be arranged if vowels alwaysoccupy the odd places.

Hint: [ Ans: 4324321=2424=576 ]Q.2121/1 Numbers of words which can be formed using all the letters of the word "AKSHI", if each word begins

with vowel or terminates in vowel. [ Ans: 224+224-26 = 84 ]Q.2222/1 A letter lock consists of three rings each marked with 10 different letters. Find the number of ways in

which it is possible to make an unsuccessful attempts to open the lock. [Ans: 103 - 1 = 999]Q.2323/1 How many 10 digit numbers can be made with odd digits so that no two consecutive digits are same.

[Ans: 549]Q.24 In how many ways can the letters of the word "CINEMA" be arranged so that the order of vowels do

not change. [Ans: 63!! = 120]

Q.2525/1 How many natural numbers are there with the property that they can be expressed as the sum of thecubes of two natural numbers in two different ways . [Ans. Infinitely many]

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Target IIT JEE 2010

Q.11/2 How many of the 900 three digit numbers have at least one even digit?(A*) 775 (B) 875 (C) 450 (D) 750

[Sol. There are 900 three digit numbers and there are five odd digits. Thus, there are 53 = 125 three digitnumbers comprised of only odd digits. The other 900 125 = 775 three digit numbers must contain atleast one even digit. ] [12 & 13 05-3-2006]

Q.22/2 The number of natual numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digitsis(A) 4048 (B*) 4464 (C) 4518 (D) 4536

ORWhat can you say about the number of even numbers under the same constraints?

[Hint: Total All four different = 7899109 3 OR

Ans. 2204 ; all 4 digit even number number of 4 digit even numbers with different digit ]

Q.33/2 The number of different seven digit numbers that can be written using only three digits1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is(A*) 672 (B) 640 (C) 512 (D) none

[Hint: Two blocks for filling 2 can be selected in 7C2 ways and the digit 25

27 2C

can be filled only in one way other 5 blocks can be filled in 25 ways.]

Q.44/2 Out of seven consonants and four vowels, the number of words of six letters, formed by taking fourconsonants and two vowels is (Assume that each ordered group of letter is a word):(A) 210 (B) 462 (C*) 151200 (D) 332640

[Hint : 7C4 4C2 . 6 ! = 151200 ]Q.55/2 All possible three digits even numbers which can be formed with the condition that if 5 is one of the digit,

then 7 is the next digit is :(A) 5 (B) 325 (C) 345 (D*) 365

[Hint : 5 + 8.9.5 = 365; 1st place is five + where 1st place is not five 75 |5

+ 5|

9|

8| ]

Q.66/2 For some natural N , the number of positive integral ' x ' satisfying the equation ,1 ! + 2 ! + 3 ! + ...... + (x !) = (N)2 is :

(A) none (B) one (C*) two (D) infinite[Hint: x = 1 & x = 3 ]Q.77/2 The number of six digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 & 7 so that digits do

not repeat and the terminal digits are even is :(A) 144 (B) 72 (C) 288 (D*) 720

[Hint: 1 . . 3 . . 5 . . 73C2 2! 5C4 4! = 6 120 = 720 ]


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Q.8 In a certain strange language, words are written with letters from the following six-letter alphabet :A, G, K, N, R, U. Each word consists of six letters and none of the letters repeat. Each combination ofthese six letters is a word in this language. The word "KANGUR" remains in the dictionary at,(A*) 248th (B) 247th (C) 246th (D) 253rd

[Sol. beginning with A or G = 240 [18-12-2005, 13th]beginning with 6247th K A N G R U248th K A N G U R ]

Q.9 Consider the five points comprising of the vertices of a square and the intersection point of its diagonals.How many triangles can be formed using these points?(A) 4 (B) 6 (C*) 8 (D) 10

[Hint: To form a triangle, 3 points out of 5 can be chosen in 5C3 = 10 ways. But of these, the three points lyingon the 2 diagonals will be collinear. So 10 2 = 8 triangles can be formed]

Q.1010/2 A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 & 5 without repetition.The total number of ways this can be done is :(A) 3125 (B) 600 (C) 240 (D*) 216

[Hint: reject 0 + reject 3 5! + 4 4! = 120 + 96 = 216 ]

Q.1111/2 Number of 9 digits numbers divisible by nine using the digits from 0 to 9 if each digit is used atmostonce is K 8 ! , then K has the value equal to ______. [Ans. K = 17]

[Hint: Case-I : reject '0' ; Case-II : reject 9Case-I: 9!; Case-II: 0, 1, 2, ......, 8; 9! 8! = 8 8!; I + II = (17)8!]

Q.1212/2 Number of natural numbers less than 1000 and divisible by 5 can be formed with the ten digits, eachdigit not occuring more than once in each number is ______ .

[Hint : single digit = 1 ;

two digit: 5 = 8 + 0 = 9 Total = 17

three digit: 5 = 8 8 = 64; 0 = 9 8 = 72 Total = 136hence Single digit + two digit + three digit = 154 Ans.]

Q.1311/3 Number of 3 digit numbers in which the digit at hundreath's place is greater than the other two digit is(A*) 285 (B) 281 (C) 240 (D) 204

[Sol. When all 3 are distinctzyx321 rdndst

10C3 2 (largest being at the 1st place and the= 240 remaining two can be arranged in two ways) [11th, 16-11-2008]

when y = z10C2 1 (largest on the 1st place and remaining two being= 45 equal on the 2nd and 3rd place) e.g. (211, 100)]

Total = 285 Ans. ]Q.1412/3 Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken all at a time, such that the digit

1 appearing somewhere to the left of 23 appearing to the left of 4 and5 somewhere to the left of 6, is

(e.g. 815723946 would be one such permutation)(A*) 9 7! (B) 8! (C) 5! 4! (D) 8! 4!

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Sol. Number of digits are 9 [12th, 28-09-2008] [Dpp-3]select 2 places for the digit 1 and 2 in 9C2 waysfrom the remaining 7 places select any two places for 3 and 4 in 7C2 waysand from the remaining 5 places select any two for 5 and 6 in 5C2 waysnow, the remaining 3 digits can be filled in 3! ways Total ways = 9C2 7C2 5C2 3!

= !7!2!9

!5!2!7

!3!2!5

3! = 8!9 = 8

!789 = 9 7! Ans. ]

Q.15 Number of odd integers between 1000 and 8000 which have none of their digits repeated, is(A) 1014 (B) 810 (C) 690 (D*) 1736

[Sol. Let the last place is 9

1/3/5/7/9

then we have 7 8 7 = 392 (1st place only 1 to 7 can come)If the last place has 1 | 3 | 5 | 7 thenwe have (6) (8) (7) (4) = 1344 [0, 9 or 8 can not come at units place]Total = 392 + 1344 = 1736] [11th, 25-01-2009, P-1]

Q.164/3Find the number of ways in which letters of the word VALEDICTORY be arranged so that the vowelsmay never be separated.

[ Hint: VLDCTRYY or 8! 4! = 40320 24 = 967680 Ans ] [Valediction means farewell after graduation from a college. Valedictory : to take farewell]

Q.177/3The number of ways in which 5 different books can be distributed among 10 people if each person canget at most one book is :(A) 252 (B) 105 (C) 510 (D*) 10C5.5!

[Hint: Select 5 boys in 10C5 and distribute 5 books in 5! ways hence 10C5. 5!]Q.188/2A new flag is to be designed with six vertical strips using some or all of the colours yellow, green, blue and

red. Then, the number of ways this can be done such that no two adjacent strips have the same colour is(A*) 12 81 (B) 16 192 (C) 20 125 (D) 24 216

[Hint: 1st place can be filled in 4 ways2nd place can be filled in 3 ways3rd place can be filled in 3 ways and |||ly 4th, 5th and 6th each can be filled in 3 ways.hence total ways = 4 35 = 12 81 ]

Q.194/4 5 Indian & 5 American couples meet at a party & shake hands . If no wife shakes hands with her ownhusband & no Indian wife shakes hands with a male, then the number of hand shakes that takes place inthe party is :(A) 95 (B) 110 (C*) 135 (D) 150

[Hint: 20C2 (50 + 5) = 135 ]

Q.20192/1(10/3) There are 720 permutations of the digits 1, 2, 3, 4, 5, 6. Suppose these permutations are arrangedfrom smallest to largest numerical values, beginning from 1 2 3 4 5 6 and ending with 6 5 4 3 2 1.(a) What number falls on the 124th position?(b) What is the position of the number 321546? [Ans. (a) 213564, (b) 267th]

[Sol. (a) digits 1, 2, 3, 4, 5, 6starting with 1, number of numbers = 120

starting with 2, 1, 3, 4 number of numbers = 2123rd

finally 124th is = 213564(b) N = 321546

number of numbers beginning with 1 = 120number of numbers beginning with 2 = 120starting with 31 .............................. = 24starting with 3214 .......................... = 2finally = 1hence N has 267th position ]

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Target IIT JEE 2010

Q.15/3 How many numbers between 400 and 1000 (both exclusive) can be made with the digits 2,3,4,5,6,0 if(a) repetition of digits not allowed.(b) repetition of digits is allowed.

[ Hint: (a) 354 = 60 ; (b) 366 = 108 1 = 107 ]

Q.29/3 The 9 horizontal and 9 vertical lines on an 8 8 chessboard form 'r' rectangles and 's' squares. The ratio

rs

in its lowest terms is

(A) 61

(B*) 10817

(C) 274

(D) none

[Sol. no. of squares are [12 & 13th test (29-10-2005)]

S = 12 + 22 + 32 + ........ + 82 = 6)17)(9(8

= 204

no. of rectangles r = 9C2 9C2 = 1296

hence rs

= 1296204

= 32451

= 10817

]

Q.311/3 A student has to answer 10 out of 13 questions in an examination . The number of ways in which he cananswer if he must answer atleast 3 of the first five questions is :(A*) 276 (B) 267 (C) 80 (D) 1200

[Hint : 13C10 number of ways in which he can reject 3 questions from the first five or 13C10 5C3 = 286 10 = 276 or 5C3 . 8C7 + 5C4 . 8C6 + 5C5 . 8C5 = 276 ][ Note that 5C3 . 10C7 is wrong ( cases repeat]

Q.412/3 The number of three digit numbers having only two consecutive digits identical is :(A) 153 (B*) 162 (C) 180 (D) 161

[Hint : when two consecutive digits are 11, 22, etc = 9 . 9 = 81 when two consecutive digits are 0 0 = 9 when two consecutive digits are 11, 22, 33, ... = 9 . 8 = 72 Total ]

Q.51/4 A telegraph has x arms & each arm is capable of (x 1) distinct positions, including the position of rest.The total number of signals that can be made is ______ . [ Ans. (x 1)x 1 ]

Q.62/4 The interior angles of a regular polygon measure 150 each . The number of diagonals of the polygon is(A) 35 (B) 44 (C*) 54 (D) 78

[Hint : exterior angle = 30

Hence number of sides = 30360

= 12

number of diagonals = 2)312(12

= 54 ]

[Note that sum of all exterior angles of a polygon = 2 and sum of all the interior angles of a polygon =(2n-4)2

]


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Q.73/4 Number of different natural numbers which are smaller than two hundred million & using only the digits1 or 2 is :(A* ) (3) . 28 2 (B) (3) . 28 1 (C) 2 (29 1) (D) none

[Hint: Two hundred million = 2 x 108; (21+ 22 +23+24+25+26+27+28) + 28 = 766 ]

Q.85/4 The number of n digit numbers which consists of the digits 1 & 2 only if each digit is to be used atleastonce, is equal to 510 then n is equal to:(A) 7 (B) 8 (C*) 9 (D) 10

[Hint: (2 x 2 x .............2) n times-(when 1 or 2 is there at all the n places] [ Ans. 2n 2 ]

Q.96/4 Number of six digit numbers which have 3 digits even & 3 digits odd, if each digit is to be used atmostonce is ______ . [ Ans. 64800 ][ when 0 is included (4C2 . 5C3 . 5 . 5 !) and when 0 is excluded (4C3 . 5C3 . 6 !) ]

[Hint: alternatively, 5C3 5C3 6! since all digits 0, 1, 2, .........8, 9 are equally likely at all places

required number = 5

35

3 610

C C ! 9 digits

or required number of ways = 5C3 . 5C3 . 6 ! 4C2 . 5C3 . (6! 5!) ]

Q.1016/9 Find the number of 10 digit numbers using the digits 0, 1, 2, ....... 9 without repetition. How many ofthese are divisible by 4.

[Sol. Digit 0, 1, 2,........8, 9For a number to be divisible by 4 the number formed by last two digits must be divisible by 4 and can be

04, 08, 12, ........, 96 ; Total of such numbers = 24Out of these 44 and 88 are to be rejected. (as repetition is not allowed)Hence accepted number of cases = 22Out of these number of cases with '0' always included are 04, 08, 20, 40, 60, 80 (six)no. of such numbers with these as last two digits = 6 8! ...(1)

e.g. [ 04 ]no. of other numbers = 16 7 7! = 14 8! ....(2)

e.g. [ 16 ] Total number = 6 8! + 14 8! = (20) 8! Ans. ]

Q.1114/8 There are counters available in x different colours. The counters are all alike except for the colour. Thetotal number of arrangements consisting of y counters, assuming sufficient number of counters of eachcolour, if no arrangement consists of all counters of the same colour is :(A*) xy x (B) xy y (C) yx x (D) yx y

Q.128/418 points are indicated on the perimeter of a triangle ABC (see figure).How many triangles are there with vertices at these points?(A) 331 (B) 408(C) 710 (D*) 711

[Hint: 18C3 3 7C3 = 816 105 = 711 ] [08-01-2005, 12th]Q.139/4An English school and a Vernacular school are both under one superintendent. Suppose that the

superintendentship, the four teachership of English and Vernacular school each, are vacant, if there bealtogether 11 candidates for the appointments, 3 of whom apply exclusively for the superintendentshipand 2 exclusively for the appointment in the English school, the number of ways in which the differentappointments can be disposed of is :(A) 4320 (B) 268 (C) 1080 (D*) 25920

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[Hint : 3C1 . 4C2 . 2 ! . 6 ! ]

Q.1410/4 A committee of 5 is to be chosen from a group of 9 people. Number of ways in which it can be formedif two particular persons either serve together or not at all and two other particular persons refuse toserve with each other, is(A*) 41 (B) 36 (C) 47 (D) 76

[Sol. 9 other5

separatedD/CAB

AB included 7C3 5C1 = 30 (7C3 denotes any 3 from (CD and 5 others no. of ways when CD)AB excluded 7C5 5C3 = 11 is taken and one 3 from remaining five)

41 ] [18-12-2005, 12th + 13th]

Q.1511/4 A question paper on mathematics consists of twelve questions divided into three parts A, B and C,each containing four questions. In how many ways can an examinee answer five questions, selectingatleast one from each part.(A*) 624 (B) 208 (C) 1248 (D) 2304

[Hint: 3 (4C2 . 4C2 . 4C1) + 3 (4C1 . 4C1 . 4C3) = 432 + 194 = 624 ]Alternative : Total [no of ways in which he does not select any question from any one section]

12C5 3 8C5 ; Note that 4C1 . 4C1 . 4C1 . 9C2 is wrong think !]Q.1612/4 If m denotes the number of 5 digit numbers if each successive digits are in their descending order of

magnitude and n is the corresponding figure, when the digits are in their ascending order of magnitudethen (m n) has the value(A) 10C4 (B*) 9C5 (C) 10C3 (D) 9C3

[Hint: m n = 10C5 9C5 = 252 125 = 126 = 9C5 or 9C4Alternatively: 9C5 + 9C4 9C5 = 9C4 = 9C5 ]Q.171/5There are m points on a straight line AB & n points on the line AC none of them being the point A.

Triangles are formed with these points as vertices, when(i) A is excluded (ii) A is included. The ratio of number of triangles in the two cases is:

(A*) m nm n

2 (B) m nm n

21

(C) 2nm2nm

(D) m n

m n( )

( ) ( )

1

1 1

[Hint:m C n C

m C n C mn

n m

n m. .

. .2 2

2 2

]

Q.183/5In a certain algebraical exercise book there are 4 examples on arithmetical progressions, 5 examples onpermutation-combination and 6 examples on binomial theorem . Number of ways a teacher can selectfor his pupils atleast one but not more than 2 examples from each of these sets, is ______.

[ Hint: ( 4C1 + 4C2) ( 5C1 + 5C2) ( 6C1 + 6C2) ] [ Ans. 3150 ][Alternatively : add one dummy exercies in each and compute 5C2 6C2 7C2 ]

Q.191/9n

rn

rn

rr

n CC C

10

1

is equal to :

(A) n n

n( )( )

12 1 (B)

n12

(C) n n( )1

2(D*)

n2

Q.208/5The number of 5 digit numbers such that the sum of their digits is even is :(A) 50000 (B*) 45000 (C) 60000 (D) none

[Hint:2109 4 ]

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Target IIT JEE 2010

Q.1 Number of ways in which 8 people can be arranged in a line if A and B must be next each other and Cmust be somewhere behind D, is equal to(A) 10080 (B*) 5040 (C) 5050 (D) 10100

[Sol.DC

AB [12th 11-05-2008]

AB and 6 other is 7! but A and B can be arranged in 2! ways Total ways = 7! 2!when C is behind D required number of ways = !2

!2!7 = 5040 ways Ans.]

Q.22/5 Number of ways in which 7 green bottles and 8 blue bottles can be arranged in a row if exactly 1 pair ofgreen bottles is side by side, is (Assume all bottles to be alike except for the colour).(A) 84 (B) 360 (C*) 504 (D) none

[Sol. G G G G G G G / B B B B B B B B [12th & 13th 03-12-2006]one gap out of nine can be taken in 9C1 waysnow green remaining 5

gaps remaining 85 gaps for remaining 5 green can be selected in 8C5

Hence Total ways 9C1 8C5 = 9 321678

= 72 7 = 504 Ans.

Alternatively: 9C6 6 = 504 Ans. (think ! how)(6 bottle in 6 gaps and the remaining in any one of these 6 gaps)]

Q.34/5 The kindergarten teacher has 25 kids in her class . She takes 5 of them at a time, to zoological garden asoften as she can, without taking the same 5 kids more than once. Then the number of visits, the teachermakes to the garden exceeds that of a kid by :(A) 25C5 24C5 (B*) 24C5 (C) 24C4 (D) none

[Hint: Number of trips which exceeds when one kid is never included 25C5 24C4 = 24C5 ]

Q.46/5 Seven different coins are to be divided amongst three persons . If no two of the persons receive thesame number of coins but each receives atleast one coin & none is left over, then the number of ways inwhich the division may be made is(A) 420 (B*) 630 (C) 710 (D) none

[Hint: 1, 2, 4 groups] [Ans. 71 2 4

!! ! !

3 ! ]

Q.57/5 Let there be 9 fixed points on the circumference of a circle. Each of these points is joined to every oneof the remaining 8 points by a straight line and the points are so positioned on the circumference thatatmost 2 straight lines meet in any interior point of the circle. The number of such interior intersectionpoints is :(A*) 126 (B) 351 (C) 756 (D) none of these

[Hint : Any interior intersection point corresponds to 4 of the fixed points , namely the 4 end points of theintersecting segments. Conversely, any 4 labled points determine a quadrilateral, the diagonals of whichintersect once within the circle . Thus the answer is A ]


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Q.610/5 The number of ways in which 8 distinguishable apples can be distributed among 3 boys such that everyboy should get atleast 1 apple & atmost 4 apples is K 7P3 where K has the value equal to(A) 14 (B) 66 (C) 44 (D*) 22

Q.711/5 A women has 11 close friends. Find the number of ways in which she can invite 5 of them to dinner, if twoparticular of them are not on speaking terms & will not attend together. [Ans. 378]

[Hint: 11C5 9C3 = 9C5 + 2 9C4 = 3 9C4 = 378 ]

5 to be

selected 9C3 + 9C4 + 9C4 or 11C5 9C3 ]

Q.812/5 A rack has 5 different pairs of shoes. The number of ways in which 4 shoes can be chosen from it , sothat there will be no complete pair is :(A) 1920 (B) 200 (C) 110 (D*) 80

[Hint: 5C4 . 24 or 10 8 6 4

4. . .

! = 80 ]

Q.93/6 There are 10 seats in a double decker bus, 6 in the lower deck and 4 on the upper deck. Ten passengersboard the bus, of them 3 refuse to go to the upper deck and 2 insist on going up. The number of waysin which the passengers can be accommodated is _____. (Assume all seats to be duly numbered)

[ Ans. 4C2. 2! 6C3. 3! 5! or 172800]Q.104/6Find the number of permutations of the word "AUROBIND" in which vowels appear in an alphabetical

order. [Ans. 8C4 4 !]

[Hint : A, I, O, U treat them alike. Now find the arrangement of 8 letters in which 4 alike and 4 different = 84

!!]

Q.115/6 The greatest possible number of points of intersection of 9 different straight lines & 9 different circles ina plane is(A) 117 (B) 153 (C*) 270 (D) none

[Hint: 9C2 . 1 + 9C1 . 9C1 . 2 + 9C2 . 2 = 270 ]

Q.126/6An old man while dialing a 7 digit telephone number remembers that the first four digits consists of one1's, one 2's and two 3's. He also remembers that the fifth digit is either a 4 or 5 while has no memorisingof the sixth digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number ofdistinct trials he has to try to make sure that he dials the correct telephone number, is(A) 360 (B*) 240 (C) 216 (D) none

[Hint:

!2!4

placefifthforways2

place6ways10

th

place7way1

th th73321

x7 = 9 x6x6 can take 0 to 9

= 240 ] [12 & 13th test (09-10-2005)]Q.137/6If as many more words as possible be formed out of the letters of the word "DOGMATIC" then the

number of words in which the relative order of vowels and consonants remain unchanged is ______.

[Hint: IA0 3! 5! 1 = 719 ]

Q.148/6Number of ways in which 7 people can occupy six seats, 3 seats on each side in a first class railwaycompartment if two specified persons are to be always included and occupy adjacent seats on the sameside, is 5 ! (k) then k has the value equal to :(A) 2 (B) 4 (C*) 8 (D) none

[Hint: including the two specified people, 4 others can be selected in 5C4 ways . The two adjacent seats can betaken in 4 ways and the two specified people can be arranged in 2 ! ways, remaining 4 people can bearranged in 4 ! ways . C4 . 4 . 2 ! 4 ! = 5 ! 8 = 8 . 5 ! C ]

[13]

Q.159/6Number of ways in which 9 different toys be distributed among 4 children belonging to different agegroups in such a way that distribution among the 3 elder children is even and the youngest one is toreceive one toy more, is :

(A) 58

2! (B) 92!

(C*) 9

3 2 3!

! ! (D) none

[Hint : distribution 2, 2, 2 and 3 to the youngest . Now 3 toys for the youngest can be selected in 9C3 ways,remaining 6 toys can be divided into three equal groups in

6

2 33!

! . ! way and can be distributed in 3 ! ways 9C3 .

62 3

!!

= 9

3 2 3!

! ! ]

Q.1610/6 In an election three districts are to be canvassed by 2, 3 & 5 men respectively . If 10 men volunteer, thenumber of ways they can be alloted to the different districts is :

(A*) 102 3 5

!! ! !

(B) 10

2 5!

! !(C)

102 52

!( !) !

(D) 10

2 3 52!

( !) ! ![Hint : number of groups of 2, 3, 5 =

102 3 5

!! ! !

& can be deputed only in one way ]

Q.1711/6 Let Pn denotes the number of ways in which three people can be selected out of ' n ' people sitting ina row , if no two of them are consecutive.If , Pn + 1 Pn = 15 then the value of 'n' is :(A) 7 (B*) 8 (C) 9 (D) 10

[Hint : Pn = n 2C3 ; Pn + 1 = n 1C3 ; Hence n 1C3 - n 2C3 = 15n 2C3 + n 2C2 n 2C3 = 15 or n 2C2 = 15 n = 8 C ]

Q.18 The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is(A*) 26 (B) 18 (C) 31 (D) none

[Hint: Draw a Venn diagram with circles namedA : the number of integers from 1 to 100 divisible by 2 = 50B : the number of integers from 0 to 100 divisible by 3 = 33C : the number of integers from 0 to 100 divisible by 5 = 20where the overlap portions are the numbers divisible by bothA & B , B & C, etc. (i.e. by 6, 10, 15 & 30).The total number of integers not divisible is then 26. ]

Use : n (A B C) = n (A) + n (B) + n(C) n(A B) n (B C) n (C A) + n (A B C) ]

Q.19203/1(2/6) In how many different ways a grandfather along with two of his grandsons and four grand daughterscan be seated in a line for a photograph so that he is always in the middle and the two grandsons arenever adjacent to each other. [Ans. 528]

[Sol. Total number of ways they can sit = 6! no. of ways when the two grandsons are always adjacent = 4 2! 4! = 192where 4 denotes the no. of adjacent positions(2!) no. of ways in which two sons can be seatedand 4! no. of ways in which the daughter can be seated in the remaining places. required no. of ways = 720 192 = 528 Ans ]

Q.209/5A forecast is to be made of the results of five cricket matches, each of which can be win, a draw or a lossfor Indian team. Find(i) the number of different possible forecasts(ii) the number of forecasts containing 0, 1, 2, 3, 4 and 5 errors respectively

[ Ans: 35 = 243 ; 1, 10, 40, 80, 80, 32 ][Hint: N(0) = 1 ; N(1) = 2.5C4 ; N(2) 22. 5C3 ; N(3) 23. 5C2 ; N(4) = 24. 5C1 ; N(5) = 25 ]

[15]


Target IIT JEE 2010

Q.14/9 There are six periods in each working day of a school. Number of ways in which 5 subjects can bearranged if each subject is allotted at least one period and no period remains vacant is(A) 210 (B*) 1800 (C) 360 (D) 3600

[Hint: 6C2 5C1 4! = 1800 S1 S2 S3 S4 S5 note that at least one of the subject has to be repeated [13th Test (5-12-2004)]two periods in which one subject is to be repeated 5C1 4! one subject ]

Q.21/7 There are 10 red balls of different shades & 9 green balls of identical shades. Then the number ofarranging them in a row so that no two green balls are together is(A) (10 !) . 11P9 (B*) (10 !) . 11C9 (C) 10 ! (D) 10 ! 9 !

Q.32/7 Number of ways in which n distinct objects can be kept into two identical boxes so that no box remainsempty, is

[Hint: Consider the boxes to be different for a moment. T1 can be kept in either of the boxes in 2 ways,silmilarly for all other things Total ways = 2nbut this includes when all the things are in B1 or B2 number of ways = 2n 2Since the boxes are identical 2 2

2

n = 2n 1 1 ]

Q.43/7 A shelf contains 20 different books of which 4 are in single volume and the others form sets of 8, 5 and3 volumes respectively. Number of ways in which the books may be arranged on the shelf, if thevolumes of each set are together and in their due order is

(A) !3!5!8!20

(B) 7! (C*) 8! (D) 7 . 8!

[Hint : Volume of each set may be in due order in two ways, either from left to right or from right to left. Nowwe haveD1 D2 D3 D4 , = 7! 2! . 2! . 2! = 8! ]

Q.59/7 In a certain college at the B.Sc. examination, 3 candidates obtained first class honours in each of thefollowing subjects: Physics, Chemistry and Maths, no candidates obtaining honours in more than one subject;Number of ways in which 9 scholarships of different value be awarded to the 9 candidates if due regard isto be paid only to the places obtained by candidates in any one subject is _____. [ Ans: 1680 ]

[ Hint : 3 candidates in P; 3 in C and 3 in M. Now 9 scholarships can be divided into three groups in

!3.)!3(!93

ways and distributed to P, C, M in 3)!3(

!9 ways ]

Q.65/7 Number of rectangles in the grid shown which are not squares is

(A*) 160 (B) 162 (C) 170 (D) 185[Sol. Total = 7C2 5C2 = 210 number of squares [13th, 19-10-2008]

number of squares = units4units3units2units1

381524 = 50

required number = 210 50 = 160 Ans. ]


[16]

Q.76/7 All the five digits number in which each successive digit exceeds its predecessor are arranged in theincreasing order of their magnitude. The 97th number in the list does not contain the digit(A) 4 (B*) 5 (C) 7 (D) 8

[Sol. All the possible number are 9C5 (none containing the digit 0) = 126 [12th test (29-10-2005)]Total numbers starting with 1 = 8C4 = 70(using 2, 3, 4, 5, 6, 7, 8, 9)Total starting with 23 = 6C3 = 20(4, 5, 6, 7, 8, 9)Total starting with 245 = 4C2 = 6(6, 7, 8, 9)97th number = ]

Q.87/7 The number of combination of 16 things, 8 of which are alike and the rest different, taken 8 at a time is______. [Ans. 256]

[Hint : A A A A A A A A D1 D2 ......... D8 8C0 + 8C1 + 8C2 + ...... + 8C8 = 256 ]Q.98/7 The number of different ways in which five 'dashes' and eight 'dots' can be arranged, using only seven of

these 13 'dashes' & 'dots' is :(A) 1287 (B) 119 (C*) 120 (D) 1235520

[Hint: 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 ]Q.1010/7 There are n identical red balls & m identical green balls . The number of different linear arrangements

consisting of "n red balls but not necessarily all the green balls" is xCy then(A) x = m + n , y = m (B*) x = m + n + 1 , y = m(C) x = m + n + 1 , y = m + 1 (D) x = m + n , y = n

[Hint: Put one more red ball & find the arrangement of n + 1 red and m green balls = m + n + 1Cm ]Q.116/8 A gentleman invites a party of m + n (m n) friends to a dinner & places m at one table T1 and n at

another table T2 , the table being round . If not all people shall have the same neighbour in any twoarrangement, then the number of ways in which he can arrange the guests, is

(A*) ( ) !m n

mn

4(B) 1

2 ( ) !m n

mn (C) 2 ( ) !m n

mn (D) none

[Hint: 2

)!1n(.2

)!1m(!n!m)!nm(

(A) ]

Q.1211/7 Consider a determinant of order 3 all whose entries are either 0 or 1. Five of these entries are 1 andfour of them are '0'. Also aij = aji 1 i, j 3. Find the number of such determinants.

[Ans. 12][Sol. We have aij = aji a12 = a21 = a (say) [11th, 28-11-2009, P-2] [Dpp, p&c]

a31 = a13 = b and a23 = a32

cbcaba

1, 1, 1, 1, 1 and 0, 0, 0, 0Case-I :If diagonal elements are (1, 1, 1 )

Conjugate element are, (0, 0), (0, 0), (1, 1) as shown. 110110001

This can be done in !2!3

= 3 ways

[17]

Case-II :If diagonal has (1, 0, 0) then the conjugate elements are(1, 1), (1, 1), (0, 0).

Now conjugate elements can be arranged in 2!3 = 3 ways

00

1

and diagonal elements can be filled in 2!3 = 3 ways

Hence 3 3 = 9 ways Total determinant = 3 + 9 = 12 Ans.]

Q.131/8 Number of different words that can be formed using all the letters of the word "DEEPMALA" if twovowels are together and the other two are also together but separated from the first two is(A) 960 (B) 1200 (C) 2160 (D*) 1440

[Hint : | D | P | M | L | can be arranged in 4 ! ways & the two gaps out of 5 gaps can be selected in 5C2 ways.

two A's and two E's can be arranged in !2!2!4

= 6 ways.

4 ! 5C2 !2!2!4

= 1440 ]

Q.14 A four digit number is called a doublet if any of its digit is the same as only one neighbour. For example,1221 is a doublet but 1222 is not. Number of such doublets are(A) 2259 (B*) 2268 (C) 2277 (D) 2349

[Sol. Case-1: abbc [13th, 14-02-2009] [Dpp, P&C] doneWe can choose a in 9 ways, b in 9 ways, and c in 9 ways. Hence, there are 93 possibilities for this caseCase-2: bbacThere are 9 possibilities for b, and 9 possibilities for each a and c. Once again, we have that there are 93possibilities for this caseCase-3: acbbOnce again, there are 9 possibilities for a and c, and 9 possibilities for b. Hence , there are 93 possibilitiesfor this caseCase-4: aabbWe have 9 possibilities for a and 9 possibilities for b. We thus have 81 possibilities for this case.We can now sum up our answers, to get a total of 3 93 + 81 = 2268 total ways Ans.

Alternatively: Total four digit numbers = 9000let us compute the number which are not doublet(a) when no two adjacent digits are same

= 9 9 9 9 = 6561(b) 3 digits in a row are same at the end of the number

if they are non zero then we have= 9 8 = 72

if they are same then = 9total for (b) = 81

(c) 3 digits at the beginning of the numbers are same, then we have

= 9 9 = 81(d) All four digits are same = 9 Number of doublets = 9000 [6561 + 81 + 81 + 9] = 2268 Ans. ]

[18]

Q.153/8In a unique hockey series between India & Pakistan, they decide to play on till a team wins 5 matches.The number of ways in which the series can be won by India, if no match ends in a draw is :(A*) 126 (B) 252 (C) 225 (D) none

[Hint : India wins exactly in 5 matches looses in none 5C0 waysIndia wins exactly in 6 matches wins the 6th and looses anyone in the 1st five 5C1 ways and so on . 5C0 + 5C1 + 6C2 + 7C3 + 8C4 = 126

Alternatively : Total number of ways in which Series can be won by India or Pakistan = 10C5

required number of ways = 10

5

2C = 126 ]

Q.1612/9 Sameer has to make a telephone call to his friend Harish, Unfortunately he does not remember the 7 digitphone number. But he remembers that the first three digits are 635 or 674, the number is odd and there isexactly one 9 in the number. The maximum number of trials that Sameer has to make to be successful is(A) 10,000 (B*) 3402 (C) 3200 (D) 5000

[Sol. There are 2 ways of filling the first 3 digits, either 635 or 674. Of the remaining 4 digits, one has to be 9and the last has to be odd. If the last digit is 9 then there are 9 ways of filling each of the remaining 3digits. thus the total number of phone numbers that can be formed are 2 93 = 1458.If the last digit is not 9, then there are only 4 ways of filling the last digit. (one of 1, 3, 5 and 7). The 9could occur in any of the 3 remaining places and the remaining 2 places can be filled in 92 ways. Thus thetotal number of such numbers is : 2 4 3 92 = 1944 1944 + 1458 = 3402 Ans ]

Q.176/9A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. Ifinternal arrangement inside the car does not matter then the number of ways in which they can travel, is(A) 91 (B) 182 (C*) 126 (D) 3920

[Sol. They can sit in groups of either 5 and 3 or 4 and 4

required number = 1!3!5

!8

+

!2!4!4!2!8

= 126 ] [12th test (05-06-2005)]

Q.183/10 One hundred management students who read at least one of the three business magazines are surveyedto study the readership pattern. It is found that 80 read Business India, 50 read Business world, and 30read Business Today. Five students read all the three magazines. How many read exactly two magazines?(A*) 50 (B) 10 (C) 95 (D) 65

[Hint: n (A B C) = 100n (A) = 80 ; n (B) = 50 ; n (C) = 30

now n (A B C) = )A(n )BA(n + n (A B C) 100 = 80 + 50 + 30 )BA(n + 5 )BA(n = 65

now n (E2) = )BA(n 3n(A B C) = 65 15 = 50]Q.1913/9 Six people are going to sit in a row on a bench. A and B are adjacent, C does not want to sit adjacent

toD.EandFcansitanywhere.Numberofwaysinwhichthesesixpeoplecanbeseated,is(A) 200 (B*) 144 (C) 120 (D) 56

[Hint: ; C and D separated ; E and F any where [11-12-2005, 11th (PQRS)] and E, F can be seated in 3! 2!

no. of gaps are 4 | | E | F |C D can be seated in 4C2 2!Total ways 3! 2! 4C2 2! = 144 Ans. ]

[19]

MATCH THE COLUMN:Q.2026/10 Column-I Column-II

(A) Number of increasing permutations of m symbols are there from the n set (P) nmnumbers {a1, a2, , an} where the order among the numbers is given bya1 < a2 < a3 < an1 < an is

(B) There are m men and n monkeys. Number of ways in which every monkey (Q) mCnhas a master, if a man can have any number of monkeys

(C) Number of ways in which n red balls and (m 1) green balls can be arranged (R) nCmin a line, so that no two red balls are together, is(balls of the same colour are alike)

(D) Number of ways in which 'm' different toys can be distributed in 'n' children (S) mnif every child may receive any number of toys, is

[Ans. (A) R; (B) S; (C) Q; (D) P] [11th pqrs & J (24-11-2006)][Sol. (A) From n elements select m in nCm ways and can be arranged only in one way.

(B) 1st working can be given to M1 M2 M3........ Mm 'm' menany one of m must k1 k2 k3.......kn 'n' monkey|||ly 2nd, 3rd , ....., nth monkey in m waytotal ways = mn

(C) Number of gaps = mselect n gaps in mCn ways for n red balls total ways = mCn

(D) 1st toy in n ways2nd toy in n waysand so on. Total ways = nm ]

[21]


Target IIT JEE 2010

Q.14/8 Number of cyphers at the end of 2002C1001 is(A) 0 (B*) 1 (C) 2 (D) 200

[Hint: 2002C1001 = )!1001()!1001()!2002(

no. of zeros in (2002)! are400 + 80 + 16 + 3 = 499

no. of zeroes in (1001 !)2 = 2(200 + 40 + 8 + 1) = 498

Hence no. of zeroes is 2)!1001()!2002(

= 1 ] [12th (25-9-2005)]

Q.25/8 Three vertices of a convex n sided polygon are selected. If the number of triangles that can be constructedsuch that none of the sides of the triangle is also the side of the polygon is 30, then the polygon is a(A) Heptagon (B) Octagon (C*) Nonagon (D) Decagon

[Sol. nC3 [n + n(n 4)] = 30n = 9 satisfies it (C) ] [11th, 13-01-2008]

Q.31/10 Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on one circle. Then the maximumnumber of circles that can be drawn so that each contains atleast three of the given points is :(A) 216 (B*) 156 (C) 172 (D) none

[Hint: 5C2 . 6C1 + 6C2 5C1 + 6C3 + 1 = 156

alternatively 11C3 5C3 + 1]Q.4 Number of 5 digit numbers divisible by 25 that can be formed using only the digits

1, 2, 3, 4, 5 & 0 taken five at a time is(A) 2 (B) 32 (C*) 42 (D) 52

[ Hint : 1, 2, 3, 4, 5, 0A number divisible by 25 if the last two digits are 25 or 50(i) Now, if 5 is not taken then number of numbers = 0(ii) If 0 is not taken = 6

(iii) if 2 is not taken then = 6 and(iv) If 1 | 3 | 4 is rejected

then in each case we have, = 6

= 4 2 . 2 1

if 3, 4, or 1 is not taken then 10 number for each case.Total = 30

Hence total = 30 + 6 + 6 = 42 ]


[22]

Q.58/8 There are 12 guests at a dinner party. Supposing that the master and mistress of the house have fixedseats opposite one another, and that there are two specified guests who must always, be placed next toone another ; the number of ways in which the company can be placed, is:(A*) 20 . 10 ! (B) 22 . 10 ! (C) 44 . 10 ! (D) none

[Hint: 6 places on either sides G1G2 will have 5 places each on either side and can be seated in 2 ways 10 2! 10! Ans ]

Q.69/8 Let Pn denotes the number of ways of selecting 3 people out of 'n' sitting in a row, if no two of themare consecutive and Qn is the corresponding figure when they are in a circle. If Pn Qn = 6, then 'n' isequal to :(A) 8 (B) 9 (C*) 10 (D) 12

[Hint : Pn = n 2C3 ; Qn = nC3 [ n + n (n 4) ] or Qn = n nC C1

42

3.

Pn Qn = 6 n = 10 ]Q.7207/1(10/8) Define a 'good word' as a sequence of letters that consists only of the letters A, B and C and in

which A never immidiately followed by B, B is never immediately followed by C, and C is never immediatelyfollowed by A. If the number of n-letter good words are 384, find the value of n. [Ans. n = 8 ]

[Sol. There are 3 choices for the first letter and two choices for each subsequent letters.Hence using fundamental principlenumber of good words = 3 2n1 = 384

2n1 = 128 = 27n = 8 Ans. ]

Q.811/8 Six married couple are sitting in a room. Find the number of ways in which 4 people can be selected so that(a) they do not form a couple (b) they form exactly one couple(c) they form at least one couple (d) they form atmost one couple

[Hint : 12C4 = 6C2 + 6C1 5C2 24 + 6C4 24 ] [Ans. 240, 240, 255, 480]Q.9 In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak after S3,

then the number of ways all the 10 speakers can give their speeches with the above restriction if theremaining seven speakers have no objection to speak at any number is(A) 10C3 (B) 10P8 (C) 10P3 (D*)

103

!

[Hint: S1 S2 S3 or S3 S1 S2 10C3 . 2 . 7 ! ]Q.102/9Let m denote the number of ways in which 4 different books are distributed among 10 persons, each

receiving none or one only and let n denote the number of ways of distribution if the books are all alike.Then :(A) m = 4n (B) n = 4m (C*) m = 24n (D) none

[Hint: m = 10C4 . 4 ! and n = 10C4 ]

Q.115/9 The number of all possible selections of one or more questions from 10 given questions, each equestionhaving an alternative is :(A) 310 (B) 210 1 (C*) 310 1 (D) 210

[Hint : 1st question can be selected in three ways and so on ]

Q.12 Number of 7 digit numbers the sum of whose digits is 61 is :(A) 12 (B) 24 (C*) 28 (D) none

[Hint: only 7, 8 and 9 can be used]

[23]

Q.139/9The number of ways of choosing a committee of 2 women & 3 men from 5 women & 6 men, if Mr. Arefuses to serve on the committee if Mr. B is a member & Mr. B can only serve, if Miss C is the memberof the committee, is(A) 60 (B) 84 (C*) 124 (D) none

[Hint : ; ;

(i) Miss C is taken(a) B included A excluded 4C1 . 4C2 = 24(b) B excluded 4C1 . 5C3 = 40

(ii) Miss C is not taken B does not comes ; 4C2 . 5C3 = 60 Total = 124

Alt. Total [A, B, C present + A,B present & C absent + B present & A, C absent]Alternatively : Case 1 : Mr. ' B ' is present ' A ' is excluded & ' C ' included

Hence number of ways = 4C2 . 4C1 = 24Case 2 : Mr. ' B ' is absent no constraintHence number of ways = 5C3 . 5C2 = 100 Total = 124 ]

Q.1410/9 Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can bedone if A must have either B or C on his right and B must have either C or D on his right is :(A) 36 (B) 12 (C) 24 (D*) 18

[Hint : when A has B or C to his right we have AB or ACwhen B has C or D to his right we have BC or BDThus we must have ABC or ABD or AC and BD

for D, E, F on a circle number of ways = 3 ! = 6for C, E, F on a circle number of ways = 3 ! = 6for , E, F the number of ways = 3 ! = 6 Total = 18 ]

Q.1511/9 There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The numberof ways in which they can be arranged in a row so that atleast one ball is separated from the balls of thesame colour, is :(A*) 6 (7 ! 4 !) (B) 7 (6 ! 4 !) (C) 8 ! 5 ! (D) none

[Hint : 92 3

!! !

number of ways when balls of the same colour are together = 9

2 3!

! ! 3 ! 4 ! = 6(7! 4!) ]

Q.161/9Product of all the even divisors of N = 1000, is(A) 32 102 (B) 64 214 (C*) 64 1018 (D) 128 106

[Sol. 1000 = 23 53 [11th, 28-11-2009, P-2] [Dpp, p&c]Number of divisors = 16 which are (20 + 21 + 22 + 23)(50 + 51 + 52 + 53)Product of all the 16 divisors is

= (56)(56)(56)(56)(26)(26)(26)(26)= 224 524

Now odd divisors are 1, 5, 25, 125and product of odd deivisors = 56 Product of all the even divisors = 224 524 = 218 518 26 = (1018)64 Ans.

Alternatively: Prodcut of all the even divisors = (24 56)(28 56) (212 56) (think!) = 224 518 = 64 1018 Ans. ]

[24]

Q.17209/1(14/7) Find the number of 4 digit numbers starting with 1 and having exactly two identical digits.[Ans. 432]

[Sol. Case-I : When the two identical digits are both unity as shown. any one place out of 3 block for unity can be taken in 3 ways and the remaining two

blocks can be filled in 9 8 ways.Total ways in this case = 3 9 8 = 216

Case-II : When the two identical digit are other than unity. ; ;

two x's can be taken in 9 ways and filled in three ways and y can be taken in 8 ways.Total ways in this case = 9 3 8 = 216

Total of both case = 432 Ans. ]Q.18 Consider the word W = MISSISSIPPI(a) If N denotes the number of different selections of 5 letters from the word W = MISSISSIPPI then N

belongs to the set(A) {15, 16, 17, 18, 19} (B) {20, 21, 22, 23, 24}(C*) {25, 26, 27, 28, 29} (D) {30, 31, 32, 33, 34}

(b) Number of ways in which the letters of the word W can be arranged if atleast one vowel is separatedfrom rest of the vowels

(A) !2!4!4161!8

(B*) !2!44161!8

(C) !2!4161!8

(D) !4165

!2!4!8

(c) If the number of arrangements of the letters of the word W if all the S's and P's are separated is (K)

!4!4!10

then K equals

(A) 56

(B*) 1 (C) 34

(D) 23

[Sol. [13th, 28-12-2008, P-1](a) S S S S ; I I I I ; P P ; M

4 alike + 1 different = 2 3 = 63 alike + 2 other alike = 2 2 = 43 alike + 2 different = 2 3 = 62 alike + 2 other alike + 1 different = 3 2 = 62 alike + 3 different = 3 = 3

25(b) Atleast one vowel is separated from the rest

= total when all vowels together

= !2!4!4!11

!2!4!8

= !2!4!4!891011 !2!4

!8

=

124

91011!2!4

!8 =

141511

!2!4!8

=

4

4165!2!4

!8 = !2!44

161!8 Ans.

[25]

(c) Separate S first| I | I | I | I | P | P | M | (8 gaps)| I | I | I | I | PP | M |

47

48 C

!4!6C

!2!4!7

; S's separated but P's may or may not be separated ;

simplifies to !4!4!7!6

= !4!4

!10 K = 1 Ans. ]

Q.19 Let A = {a, b, c, d, e, f} and B = {1, 2, 3} are two sets.Let m denotes the number of mappings which are into from A to B

n denotes the number of mappings which are injective from B to A.Find (m + n). [Ans. 309]

[Sol. m = into mappings from A to BCase-I: If exactly one element of set B is not the imageof any of the elements of set A then total number of intofunctions are

3C1 (26 2) = 3 62 = 186Case-II: If exactly two elements of set B is not the imageof any of the elements of set A then total number of intofunctions are

3C2 1 = 3and hence m = 186 + 3 = 189 [12th, 19-07-2009, P-1]

Alternatively:Total mappings = 36 = 729Surjective mappings

2 2 2 908720

!3)!2(!3!6

3

2 3 1 3602720

!3!2!3!6

1 1 4 9042720

!4!2!3!6

Total surjective mapping = 540 not surjective mappings = 729 540 = 189 = mn = number of injective mapping from B to A.Injective mapping = 6C3 3! = 20 6 = 120 = n (m + n) = 189 + 120 = 309 Ans. ]

[26]

Q.20 Let a function f is defined as f : {1, 2, 3, 4} {1, 2, 3, 4}. If f satisfy )x(ff = f (x), x {1,2,3,4}then find the number of such function. [Ans. 41]

[Sol. )x(ff = f (x) ; f (x) = y f (y) = y [12th, 23-08-2009, P-2]Case-1: range contains exactly one element

it can be done in 4C1 ways say 1 remaining 3 elements i.e. 2, 3, 4 can be mapped only in one ways total = 4C1 = 4

Case-2: range contains two elements this can this can be done in 4C2 ways say 1, 2 f (1) = 1; f (2) = 2remaining 2 elements i.e. 3 and 4 each can bemapped in 2 waysTotal = 4C2 22 = 24

Case-3: range contains 3 elements whichcan be done in 4C3 ways say 1, 2, 3 f (1) = 1; f (2) = 2 and f (3) = 3now remaining 4 can be mapped only in3 waysTotal = 4C3 3 = 12 ways

Case-4: range contains all 4 elementsf (1) = 4; f (2) = 2 ; f (3) = 3; f (4) = 4only 1 way

Total = 4 + 24 + 12 + 1 = 41 Ans. ]

[27]


Target IIT JEE 2010

Choose the correct alternative (only one is correct):Q.1 Number of ways in which four different toys and five indistinguishable marbles can be distributed between

Amar, Akbar and Anthony , if each child receives atleast one toy and one marble, is(A) 42 (B) 100 (C) 150 (D*) 216

[Sol. Toys in group 1 1 2 !3!2!2!1!1!4

= 36

Marbles OOOO = 4C2 = 6 Total ways = 36 6 = 216 ]

Q.2 A 3 digit palindrome is a 3 digit number (not starting with zero) which reads the same backwards asforwards. For example 171. The sum of all even 3 digit palindromes, is(A) 22380 (B) 25700 (C*) 22000 (D) 22400

[Sol. Number of even 3 digit palindromes = 4 10 = 40UTH

[12th & 13th 03-03-2007]4 ways to fill (hundredth and unit's place) and 10 ways to fill tenth's place

sum = )place100( th

)8642(10100 +

)placeunitsofsum(

)8642(10 +

)place10( th

)9.....321(104

= 20,000 + 200 + 1800 = 22000]Q.33/10 There are 100 different books in a shelf. Number of ways in which 3 books can be selected so that no

two of which are neighbours is(A) 100C3 98 (B) 97C3 (C) 96C3 (D*) 98C3

[11th & 13th, 16-12-2007]Q.4 A lift with 7 people stops at 10 floors. People varying from zero to seven go out at each floor. The

number of ways in which the lift can get emptied, assuming each way only differs by the numberof people leaving at each floor, is :(A) 16C6 (B) 17C7 (C*) 16C7 (D) none

[Hint : consider each floor to be a beggar. Now distribute 7 identical coins ( 7 people) in 10 beggars eachreceiving none, one or more 16C7 ]

Q.5 You are given an unlimited supply of each of the digits 1, 2, 3 or 4. Using only these four digits, youconstruct n digit numbers. Such n digit numbers will be called L E G I T I M A T E if it contains the digit1 either an even number times or not at all. Number of n digit legitimate numbers are(A) 2n + 1 (B) 2n + 1 + 2 (C) 2n + 2 + 4 (D*) 2n 1(2n + 1)

[Sol. ............ (n places) [12th test (09-10-2005)]Total numbers = 3n + nC2 3n 2 + nC4 3n 4 + .......... (nC2 indicates selection of places)

= 2

)13()13( nn = 2

1[4n + 2n] = 22n 1 + 2n 1 = 2n 1(2n + 1) ]

Q.65/10 Two classrooms A and B having capacity of 25 and (n25) seats respectively.An denotes the number ofpossible seating arrangements of room 'A', when 'n' students are to be seated in these rooms, startingfrom room 'A' which is to be filled up full to its capacity. If An An1 = 25! (49C25) then 'n' equals(A*) 50 (B) 48 (C) 49 (D) 51

[Hint: Given An = nC25 25! ; An 1 = n1C25 25!hence nC25. 25! n1C25 . 25! = 25! 49C25


[28]

or n1C25 + n1C24 n1C25 = 49C24 n 1 = 49 n = 50 n1C24 = 49C24 n = 50]

Q.79/10 Number of positive integral solutions satisfying the equation (x1 + x2 + x3) (y1 + y2) = 77, is(A) 150 (B) 270 (C*) 420 (D) 1024

[Sol. (x1 + x2 + x3) (y1 + y2) = 11 7 or 7 11 [29-01-2006, 12 & 13]In the first case (x1 + x2 + x3) = 11 and (y1 + y2) = 7, which have 10C2 6C1 solutions (using beggar)In the second case (x1 + x2 + x3) = 7 and (y1 + y2) = 11, which have 6C2 10C1 solutions (using beggar) total number of solutions = 10C2 6C1 + 6C2 10C1 = 270 + 150 = 420 Ans. ]

Q.810/10 Distinct 3 digit numbers are formed using only the digits 1, 2, 3 and 4 with each digit used at most oncein each number thus formed. The sum of all possible numbers so formed is(A*) 6660 (B) 3330 (C) 2220 (D) none

[Hint: all possible = 246(1 + 2 + 3 + 4)(1 + 10 + 102) = 6 10 111 = 6660reject 1 or 2 or 3 or 4 ] [12 & 13th test (09-10-2005)]

Q.9 There are counters available in 3 different colours (atleast four of each colour). Counters are all alikeexcept for the colour. If 'm' denotes the number of arrangements of four counters if no arrangementconsists of counters of same colour and ' n' denotes the corresponding figure when every arrangementconsists of counters of each colour, then :(A) m = 2 n (B*) 6 m = 13 n (C) 3 m = 5 n (D) 5 m = 3 n

[Hint : m = 34 3 = 78; n = 34 3 2 2 34 = 81 45 = 36Hence m

n =

7836

= 136

6 m = 13 n B ]

Q.1012/10 An ice cream parlour has ice creams in eight different varieties. Number of ways of choosing 3 icecreams taking atleast two ice creams of the same variety, is(Assume that ice creams of the same variety to be identical & available in unlimited supply)(A) 56 (B*) 64 (C) 100 (D) none

[Hint: 10C3 8C3 = 120 56 = 64 ] [Note: 10C3 = all diff. + 2 alike and 1 diff. + all alike]Q.1113/10 There are 12 books on Algebra and Calculus in our library , the books of the same subject being

different. If the number of selections each of which consists of 3 books on each topic is greatest then thenumber of books of Algebra and Calculus in the library are respectively:(A) 3 and 9 (B) 4 and 8 (C) 5 and 7 (D*) 6 and 6

[Hint: = xC3 12 xC3 ]

Q.12 Three digit numbers in which the middle one is a perfect square are formed using the digits 1 to 9 . Theirsum is :(A*) 134055 (B) 270540 (C) 170055 (D) none of these

[ Hint: Middle place 1, 4 & 9Two terminal positions 1, 2, ...... , 9

Hence total numbers = 9 . 9 . 3 = 243 (Terminal digits in 9 ways and middle one in 3 ways)For the middle place 1, 4 & 9 will come 81 times sum = 81 10 (1 + 4 + 9) AFor units place each digit from 1 to 9 will appear 27 times sum = 27 (1 + 2 + ...... + 9) BFor hundreath's place, similarly sum = 27 100 (1 + 2 + ...... + 9) C A + B + C gives the required sum ]

[29]

Q.13 A guardian with 6 wards wishes everyone of them to study either Law or Medicine or Engineering.Number of ways in which he can make up his mind with regard to the education of his wards if every oneof them be fit for any of the branches to study, and atleast one child is to be sent in each discipline is :(A) 120 (B) 216 (C) 729 (D*) 540

[Hint: Divide 6 children into groups as 123, 411 ot 222

Now total = !3!2!1!6

3! + !2!1!1!4!6

3! + !3!2!2!2!3.!6

= 360 + 90 + 90 = 540 ]

Q.1415/8 There are (p + q) different books on different topics in Mathematics. (p q)If L = The number of ways in which these books are distributed between two students X and Y suchthat X get p books and Y gets q books.

M = The number of ways in which these books are distributed between two students X and Y such thatone of them gets p books and another gets q books.

N = The number of ways in which these books are divided into two groups of p books and q booksthen,(A) L = M = N (B) L = 2M = 2N (C*) 2L = M = 2N (D) L = M = 2N

Q.15 Number of ways in which 5 A's and 6 B's can be arranged in a row which reads the same backwardsand forwards, is(A) 6 (B) 8 (C*) 10 (D) 12

[Sol. A A A A A | B B B B B BMiddle digit must be A (think !)

M

so that even number of A's and B's are availableTake AABBB on one side of M (6th place) and then their image about M in a unique way

Number of ways = !3!2!5

= 10 Ans.] [12th, 14-06-2009, P-2]

Q.16 A person writes letters to his 5 friends and addresses the corresponding envelopes. Number of ways inwhich the letters can be placed in the envelope, so that atleast two of them are in the wrong envelopes,is,(A) 1 (B) 2 (C) 118 (D*) 119

[Hint: 5! = W4andR1 + W3

andR2 + W2andR3 +

wayone

usednoneR54 + all 5 wrong

hence at least 2 in wrong = 120 1 = 119 ]Q.17 For a game in which two partners oppose two other partners, 8 men are available. If every possible pair

must play with every other pair, the number of games played is(A) 8C2 . 6C2 (B) 8C2 . 6C2 . 2 (C*) 8C4 . 3 (D) none

[Hint: For each game 4 persons are needed . Hence select 4 from 8 in 8C4 way . Now from each selection3 games can be had 8C4 . 3 = 210 ]

Q.18 The number 916238457 is an example of nine digit number which contains each of the digit 1 to 9exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits1to6donot.Numberofsuchnumbersare(A) 2268 (B*) 2520 (C) 2975 (D) 1560

[Sol. 9, 8 & 7 can be placed in 9 8 7 ways.There are only five ways to place the 6 (any where except the right most remaining slot, think !) and theorder of 1 - 5 is fixed.

[30]

Number of such numbers are 9 8 7 5 = 2520 Ans.Alternatively-1: 6 places can be selected in 9C6 ways and 6 can be placed only at 5 places except the right

most of the 6 selected. Remaining numbers i.e. 7, 8, 9 in 3! ways.Hence number of numbers 9C6 5 3! = 84 30 = 2520 Ans.

Alternatively-2:

b6

9

a5

9 !3C!4C

a = selection of 5 places for 1, 2, 3, 4, 5 & 4! for 6, 7, 8, 9 = total waysb = selection 6 places for 1, 2, 3, 4, 5, 6 & 3! for 7, 8, 9 ] [12th & 13th 07-01-2007]

Q.19 Number of functions defined from f : {1, 2, 3, 4, 5, 6} {7, 8, 9, 10} such that the sumf (1) + f (2) + f (3) + f (4) + f (5) + f (6) is odd, is

(A) 210 (B*) 211 (C) 212 (D) 212 1[Sol. Giving any random configuration for f (1) to f (5), number of functions = 45 ways (each element from 1

to 5 can be mapped in 4 ways)

Now the sum

5

1i)i(f is either odd or even

if it is odd then we have only two choices for f (6) i.e. the element 8 or 10.(as odd + even = odd).If the sum is even then also we have only choices for f (6) i.e. 7 or 9.hence the total functions = 45 2 = 211 Ans.] [12th, 24-08-2008] [TN for 11th pqrs]

Paragraph for Question Nos. 20 to 2216 players P1, P2, P3,.......P16 take part in a tennis tournament. Lower suffix player is better than anyhigher suffix player. These players are to be divided into 4 groups each comprising of 4 players and thebest from each group is selected for semifinals.

Q.20 Number of ways in which 16 players can be divided into four equal groups, is

(A*)

8

1r)1r2(

2735

(B)

8

1r)1r2(

2435

(C)

8

1r)1r2(

5235

(D)

8

1r)1r2(

635

Q.21 Number of ways in which they can be divided into 4 equal groups if the players P1, P2, P3 and P4 are indifferent groups, is :

(A) 36)!11(

(B) 72)!11(

(C*) 108)!11(

(D) 216)!11(

Q.22 Number of ways in which these 16 players can be divided into four equal groups, such that when the

best player is selected from each group, P6 is one among them, is (k) 3)!4(!12

. The value of k is :

(A) 36 (B) 24 (C) 18 (D*) 20

[Sol.(i) Number of ways = !4)!4()!16(

4 = !4!4!4!4!4

)1r2(!828

1r

8

= )24()24()24()24(567828

= )24()24()24(3529

[31]

=

8

1r)1r2(

2735

Ans. [11th, 25-01-2009, P-1]

(ii) 16P ,P ,P ,P1 2 3 4

12 others

12 others can be divided into 4 equal groups in each of 3 person 4)!3()!12(

= 6666)!11()12(

= 108)!11(

Ans.

(iii) 16P ,P ,P ,P ,P1 2 3 4 5P6P P (10)7 16

; 103

7 = !7!3

!10

0 0 0+P6

4 players

x x x x x x xP to P1 5

12 players

Now, 12444 = !3)!4(

!123

Total ways = !7!3)!10(

!3)!4()!12(

3 =

3)!4(!12

20 k = 20 Ans.]

Choose the correct alternatives (More than one are correct):Q.2315/10 The combinatorial coefficient C(n, r) is equal to

(A*) number of possible subsets of r members from a set of n distinct members.(B*) number of possible binary messages of length n with exactly r 1's.(C) number of non decreasing 2-D paths from the lattice point (0, 0) to (r, n).(D*) number of ways of selecting r things out of n different things when a particular thing is always

included plus the number of ways of selecting 'r' things out of n, when a particular thing is alwaysexcluded.

[Hint: In (C) number = n+rCr ; (D) nCr = n 1Cr 1 + n 1Cr ][18-12-2005, 12th & 13th]

Q.2417/10 There are 10 questions, each question is either True or False. Number of different sequences ofincorrect answers is also equal to(A) Number of ways in which a normal coin tossed 10 times would fall in a definite order if both Heads

and Tails are present.(B*) Number of ways in which a multiple choice question containing 10 alternatives with one or more

than one correct alternatives, can be answered.(C*) Number of ways in which it is possible to draw a sum of money with 10 coins of different

denominations taken some or all at a time.(D) Number of different selections of 10 indistinguishable things taken some or all at a time.

[Sol. 210 1; (A) 210 2; (B) 210 1; (C) 210 1; (D) 10 ] [11th & 13th, 16-12-2007]

Q.2519/10 The continued product, 2 . 6 . 10 . 14 ...... to n factors is equal to :(A) 2nCn (B*) 2nPn(C*) (n + 1) (n + 2) (n + 3) ...... (n + n) (D) none

[32]

[Hint: E = 2 6 10 14 ........ 2n = 2n(1 3 5 ....... n) = !n)!n2(

= 2nPn

also E = !n!n)nn).....(2n)(1n(

]

Q.2620/10 The Number of ways in which five different books to be distributed among 3 persons so that eachperson gets at least one book, is equal to the number of ways in which(A) 5 persons are allotted 3 different residential flats so that and each person is alloted at most one flatand no two persons are alloted the same flat.(B*) number of parallelograms (some of which may be overlapping) formed by one set of 6 parallel linesand other set of 5 parallel lines that goes in other direction.(C*) 5 different toys are to be distributed among 3 children, so that each child gets at least one toy.(D*) 3 mathematics professors are assigned five different lecturers to be delivered , so that each professorgets at least one lecturer.

[Hint: Given answer is 150 which comes in BCD in A, it is 5C3 3! = 60]

Q.2721/10 The maximum number of permutations of 2n letters in which there are only a's & b's, taken all at a timeis given by :

(A*) 2nCn (B*) 21

62

103

4 61

4 2. . ...... .nn

nn

(C*) n n n n nn

nn

11

22

33

44

2 11

2. . . ...... . (D*) 2 1 3 5 2 3 2 1n n n

n. . . ...... ( ) ( )

!

Q.2822/10 Number of ways in which 3 numbers in A.P. can be selected from 1, 2, 3, ...... n is :

(A) n

12

2

if n is even (B) n n 2

4 if n is odd

(C*) n14

2

if n is odd (D*) n n 2

4 if n is even

[Hint : n = 2m, arrange the numbers into disjoint sets1, 3, 5, .................... (2m 1) m number2, 4, 6, .....................2m m numbersno. of AP's = mC2 + mC2 ]

Q.2923/10 The combinatorial coefficient n 1Cp denotes(A) the number of ways in which n things of which p are alike and rest different can be arranged in a circle.(B*) the number of ways in which p different things can be selected out of n different thing if a particular

thing is always excluded.(C) number of ways in which n alike balls can be distributed in p different boxes so that no box remains

empty and each box can hold any number of balls.(D*) the number of ways in which (n 2) white balls and p black balls can be arranged in a line if blackballs are separated, balls are all alike except for the colour. [11th & 13th, 16-12-2007]

Q.3025/10 Which of the following statements are correct?(A*) Number of words that can be formed with 6 only of the letters of the word "CENTRIFUGAL" if

each word must contain all the vowels is 3 7!(B*) There are 15 balls of which some are white and the rest black. If the number of ways in which the

balls can be arranged in a row, is maximum then the number of white balls must be equal to 7 or 8.Assume balls of the same colour to be alike.

[33]

(C) There are 12 things, 4 alike of one kind, 5 alike and of another kind and the rest are all different. Thetotal number of combinations is 240.

(D*) Number of selections that can be made of 6 letters from the word "COMMITTEE" is 35.

[Sol. (A) 11 c7\v4/

7C2 6! = 3 7 6! = 3 7!

(B) No. of ways = )!r15(!r!15 = 15cv

r

W.....WW

r15

B......BBB

This is maximum if r = 7 or 8

(C) Total no. of combinations = 5 6 23 1 = 240 1 = 239

(D) 2 alike + 2 other alike + 2 other different = 12 alike + 2 other alike + 2 different = 3C2

4C2 = 182 alike + 4 different = 3C1

5C1 = 15All 6 different = 1

= 35 Ans. ]

Q.31 Coefficient of x2 y3 z4 in the expansion of (x + y + z)9 is equal to(A*) the number of ways in which 9 things of which 2 alike of one kind, 3 alike of

2nd kind, and 4 alike of 3rd kind can be arranged.(B) the number of ways in which 9 identical things can be distributed in 3 persons each receiving atleast

two things.(C) the number of ways in which 9 identical things can be distributed in 3 persons each receiving none

one or more.(D*) the number of ways in which 9 different books can be tied up in to three bundles one containing 2,other 3 and third containing 4 books.

[Hint: Consider (x + y + z)9 = 9Cr x9 r (y + z)r = 9Cr x9 r rCp yr p zp [x + (y + z)]9put r = 7 ; p = 4

or 9C2 7C3 4C4 = !4!3!2!9

]

Q.32 Number of ways in which the letters of the word 'B U L B U L' can be arranged in a line in a definiteorder is also equal to the(A*) number of ways in which 2 alike Apples and 4 alike Mangoes can be distributed in 3 children sothat each child receives any number of fruits.(B) Number of ways in which 6 different books can be tied up into 3 bundles, if each bundle is to haveequal number of books.(C*) coefficient of x2y2z2 in the expansion of (x + y + z)6.(D*) number of ways in which 6 different prizes can be distributed equally in three children.

[Explanation: B U L B U L; number of ways = !2!2!2!6

(A) 2 Apples can be distributed in 3 people in 4C2 way O O and 4 Mangoes in 6C2 ways O O O O

Total ways = 6C2 4C2 = !4!2!6

!2!2!4

= !2!2!2!6

(A) is correct

[34]

(B) 6 books in 3 bundles, two in each bundle = !3!2!2!2!6

(B) is incorrect(C) Tr + 1 in (x + y + z)6 is 6Cr(x + y)6 r zr

put r = 2T3 = 6C2(x + y)4 z2 = 6C2 z2 (4Cp x4 p yp) [11th pqrs & J, 21-1-2007]

put p = 2 = 6C2 z2 4C2x2y2) = 6C2 4C2 x2y2z2

= !2!2!2!6

(C) is correct

(D) 6 prizes in 3 childrents, two to each = !3!2!2!2!3!6

= !2!2!2!6

(D) is correct]

MATCH THE COLUMN:Q.3334/10 Column-I Column-II

(A) Four different movies are running in a town. Ten students go to watch (P) 11these four movies. The number of ways in which every movie is watchedby atleast one student, is(Assume each way differs only by number of students watching a movie) (Q) 36

(B) Consider 8 vertices of a regular octagon and its centre. If T denotes thenumber of triangles and S denotes the number of straight lines that canbe formed with these 9 points then the value of (T S) equals

(C) In an examination, 5 children were found to have their mobiles in their (R) 52pocket. The Invigilator fired them and took their mobiles in his possession.Towards the end of the test, Invigilator randomly returned their mobiles. Thenumber of ways in which at most two children did not get their own mobiles is (S) 60

(D) The product of the digits of 3214 is 24. The number of 4 digit naturalnumbers such that the product of their digits is 12, is

(E) The number of ways in which a mixed double tennis game can bearranged from amongst 5 married couple if no husband & wife plays (T) 84in the same game, is

[Ans. (A) T; (B) R; (C) P; (D) Q; (E) S][Sol.(A) x + y + z + t = 10 [11th, 13-01-2008]

4 movies 10 people (people as alike objects and movies as 4 different beggars)give one to each x, y, z and t andhence the number of ways = the number of non-negative integral solution of the equation

x + y + z + t = 6using Beggar method 000000

number of ways is 9C3 = 84 Ans.(B) ( 9C3 4 ) 8C2 = 52 Ans.(C) at most two children got mobile of the other children

exactly 3R and 2W + all 5 got their own mobile= 5C3 1 + 1 = 11 Ans.

(D) 12 = 22 3hence groups of 4 that work out

1, 1, 6, 2 = !2!4

= 12

[35]

1, 1, 3, 4 = !2!4

= 12

2, 2, 3, 1 = !2!4

= 12

36 Ans.(E) 5C2 . 3C2 . 2 ! = 60 Ans. ]

Subjective:Q.3428/10 A commitee of 10 members is to be formed with members chosen from the faculties of Arts, Economics,

Education, Engineering, Medicine and Science. Number of possible ways in which the facultiesrepresentation be distributed on this committee, is ________.(Assume every department contains more than 10 members). [4]

[Ans. 3003][Sol. (a) 10 members to be chosen from 6 different faculties

10

0..........000

5

(Use Beggar) ScienceF

MedicineEgEngineerinD

EducationCEconomicsBArtsA

number of ways = 15C5 = 3003 Ans. ] [11th & 13th, 16-12-2007]

Q.3532/10 On the normal chess board as shown, I1 & I2 are two insectswhich starts moving towards each other. Each insect moving withthe same constant speed . Insect I1 can move only to the right orupward along the lines while the insect I2 can move only to the left ordownward along the lines of the chess board. Find the total numberof ways thetwo insects can meet at same point during their trip.

[ Hint: (8C0 . 8C0) + (8C1 . 8C1) + .... + (8C8 . 8C8) = 16C8) = 12870 ]

Q.36186/1(30/10) 10 identical ball are distributed in 5 different boxes kept in a row and labled A, B, C, D and E.Find the number of ways in which the ball can be distributed in the boxes if no two adjacent boxesremain empty. [Ans. 771 ways]

[Sol. Case-1 : When no box remains emptyit is equivalently distributing10 coins in 5 beggar

45

00000 = 9C4 = 126

Case-2: Exactly one is empty5C1 000000

6 = 5

9C3 = 420

Case-3: Exactly two remains empty

( 5C2 two adjacent) 9C2 00000007

(10 4) 9C26 36 = 216

Case-4 : Exactly three empty. There is only 1 way to select 3 if no two adjacent

[36]

Hence 1 9C1 = 9 0 0 0 0 0 0 0 0 Total = 771 ways ]

Q.3738/10 Find the number of ways in which 12 identical coins can be distributed in 6 different purses, if not morethan 3 & not less than 1 coin goes in each purse. [Ans. 141]

[Hint: Put one coin in each purse. Now 6 coins remaining.case (i) 1 coin in each of six purses = 1

O O O O O Ocase (ii) 2 coin in each of three purses = 6C3 1 = 20

OO OO

OO

case (iii) 2 coin each in two purses and 1 coin each in 2 purses = 6C2 . 4C2 = 90

OO OO O O case (iv) 2 coins in one purse and 1 coin each in 4 purses = 6C1 . 5C4 = 30

OO O O O O Total = 140

Alternatively : coefficient of x12 in (x + x2 + x3)6 = 141 ]

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Permutations and Combinations DPP(bansal)