PERMUTATION & COMBINATION - nucleoniitjeekota.com

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PERMUTATION & COMBINATION

SyllabusPermutations and combinations

Name : ____________________________ Contact No. __________________

ARRIDE LEARNING ONLINE E-LEARNING ACADEMYA-479 indra Vihar, Kota Rajasthan 324005

Contact No. 8033545007

Page No. # 1Arride learning Online E-learning AcademyA-479 Indra Vihar, Kota Rajasthan 324005

KEY CONCEPTS

DEFINITION :

1. PERMUTATION :

Each of the arrangements in a definite order which can be made by taking some or all of a

number of things is called a PERMUTATION

2. COMBINATION :

Each of the groups or selections which can be made by taking some or all of a number of

things without reference to the order of the things in each group is called a COMBINATION

FUNDAMENTAL PRINCIPLE OF COUNTING :

If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’

different ways, then the total number of differentways of simultaneous occurrence of both

events in a definite order is m x n. This can be extended to any number of events.

RESULTS :

(i) A Useful Notation : n! = n (n – 1) (n – 2) ..............3. 2. 1; n! = n. (n – 1)! 0! = 1! = 1 ;

(2n)! = 2n . 2n. n! [1. 3. 5. 7........(2n – 1)]

Note that factorials of negative integers are not defined.

(ii) If nPr denotes the number of permutations of n different things, taking r at a time, then

nPr = n (n – 1) (n – 2) ............. (n – r + 1) = !)rn(!n

- Note that, nPn = n!

(iii) If nCr denotes the number of combinations of n different things taken r at a time, then

!rP

!)rn(!r!nC r

n

rn =

-= where r £ n ; n Î N and r Î W..

(iv) The number of ways in which (m + n) different things can be divided into two groups

containing m & n things respectively is : !n!m!)nm( +

If m = n, the groups are equal & in this

case the number of subdivision is !2!n!m!)n2(

; for in any one way it is possible to inter

change the two groups without obtaining a new distribution. However, if 2n things are to

be divided equally between two persons then the number of ways = !n!n!)n2(

.

(v) Number of ways in which (m + n + p) different things can be divided into three groups

containing m, n & p things respectively is !P!n!m!)pnm( ++, .pnm ¹¹ If m = n = p then the

number of groups = !3!n!n!n!)n3(

. However If 3n things are to be divided equally among

three people then the number of ways 3)!n(!)n3(

P E R M U T A T I O N & C O M B I N AT I O N


(vi) The number of permutations of n things taken all at a time when p of them are similar& of one type, q of them are similar & of another type, r of them are similar & third type

& the remaining n – (p + q+ r) are all dif ferent is : !r!q!p!n

.

(vii) The number of circular permutations of n different things all at a time is ; (n – 1)!. Ifclockwise & anti-clockwise circular permutations are considered to be same, then it

is 2!)1n( - .

Note : Number of circular permutations of n things when p alike and the rest different

taken all at a time distinguishing clockwise and anticlockwise arrangement is !p!)1n( -

(viii) Given n different objects , the number of ways of selecting atleast one of them is ,nC1 + nC2 + nC3 +........+ nCn = 2n – 1. This can also be stated as the total number of

combinations of n distinct things.

(ix) Total number of ways in which it is possible to make a selection by taking some or all

out of p + q + r +......things, where p are alike of one kind, q alike of a second kind, r

alike of third kind & so on is given by : (p + 1) (q + 1) (r + 1).........–1.

(x) Number of ways in which it is possible to make a selection of m + n + p = N things,

where p are alike of one kind, m alike of second kind, n alike of a third kind taken r at

a time is given by coefficient of xr in the expansion of

(1 + x + x2 +........+ xp) (1 + x + x2 +........+ xm) (1 + x + x2 +........+ xn)

Note:Remember that coefficient of xr in (1 – x)–n = n+r–1Cr (n Î N).For example the number

of ways in which a selection of four letters can be made from the letters of the word

PROPORTION is given by coefficient of x4 in (1 + x + x2 + x3) (1 + x + x2 ) (1 + x + x2 )

(1 + x) (1 + x).

(xi) Number of ways in which n distinct things can be distributed to p persons if there is no

restriction to the number of things received by men = pn

(xii) Number of ways in which n identical things may be distributed among p persons if

each person may receive none, one or more things is ; n+p–1Cn.

(xiii) (a) nCr = nCn–r ;

nC0 = nCn = 1 ; (b) nCx

= nCy Þ x = y or x + y = n

(c) nCr + nCr–1 = n+1Cr

(xiv) nCr is maximum if :

(a) 2nr = if n is even. (b) 2

1nr -= or 21nr += if n is odd.

(xv) Let N = pa. qb. rc .......where p, q, r........ are distinct primes & a, b, c....... are natural numbers then :


(a) The total numbers of divisors of N including 1 & N is = (a + 1) (b + 1) (c + 1).......

(b) The sum of these divisors is =

(p0 + p1 + p2 + ........+ pa) (q0 + q1 + q2 + ........+ qb) (r0 + r1 + r2 + ........+ rc).....

(c) Number of ways in which N can be resolved as a product of two factor is =

[ ] squareperfectaisNif1)......1c()1b()1a(21

squareperfectanotisNif)......1c()1b()1a(21

++++

+++

(d)Number of ways in which a composite number N can be resolved into two factors which

are relatively prime (or coprime) to each other is equal to 2n–1 where n is the number of

dif ferent prome factors in N.

(xvi) Grid problems and tree diagrams.

(xvii) Some times students find it difficult to decide whether a problem is on permutation or

combination or both. Based on certain words / phrases occuring in the problem we can

fairly decide its nature as per the following table :

PROBLEMS OF COMBINATIONS PROBLEMS OF PERMUTATIONS

Selections, choose Arrangementss

Distributed group is formed Standing in a line seated in a row

Committee problems on digit

geometrical problems Problems on letters from a word


PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) : Problem based, Arrangements of given objects/Selection of given object (PCAD/PCSD)

A-1. Let Pm stand for mPm. Then the expression 1. P1 + 2. P2 + 3. P3 +...... + n. Pn =(A) (n + 1) ! - 1 (B) (n + 1) ! + 1 (C) (n + 1) ! (D) none

A-2. The number of signals that can be made with 3 flags each of different colour by hoisting 1 or 2 or 3 abovethe other is:(A) 3 (B) 7 (C) 15 (D) 16

A-3. 10 different letters of an alphabet are given. Words with 5 letters are formed from these given letters. Thenthe number of words which have atleast one letter repeated is:(A) 69760 (B) 30240 (C) 99748 (D) none

A-4. The number of numbers from 1000 to 9999 (both inclusive) that do not have all 4 different digits is:(A) 4048 (B) 4464 (C) 4518 (D) 4536

Section (B) : Problem based on Selection as well as arrangement of objects/Rank of word (PCSR/PCSI)

B-1. 8 chairs are numbered from 1 to 8. 2 women & 3 men wish to occupy one chair each. First the womenchoose the chairs from amongst the chairs marked 1 to 4, then the men select the chairs from among theremaining. The number of possible arrangements is:(A) 6C3.

4C4 (B) P2. 4P3 (C) 4C3.

4P3 (D) 4P2. 6P3

B-2. How many words can be made with the letters of the word "GENIUS" if each word neither begins with Gnor ends in S is:(A) 24 (B) 240 (C) 480 (D) 504

B-3. 5 boys & 3 girls are sitting in a row of 8 seats. Number of ways in which they can be seated so that not allthe girls sit side by side is:(A) 36000 (B) 9080 (C) 3960 (D) 11600

B-4. The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which haveno digit repeated is(A) 16 × 4! (B) 1111 × 3! (C) 16 × 1111 × 3! (D) 16 × 1111 × 4!.

B-5. The number of words that can be formed by using the letters of the word ‘MATHEMATICS’ that start as wellas end with T is(A) 80720 (B) 90720 (C) 20860 (D) 37528

B-6. How many nine digit numbers can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digitsoccupy even positions?(A) 7560 (B) 180 (C) 16 (D) 60

B-7. The number of permutations that can be formed by arranging all the letters of the word ‘NINETEEN’ inwhich no two E’s occur together is

(A) !3!3!8

(B) 2

6C!3!5

´ (C) !3!5 × 6C3 (D) !5

!8 × 6C3.


B-8. Out of seven consonants & four vowels, the number of words of six letters, formed by taking four consonants& two vowels is (Assume each ordered group of letter is a word)(A) 210 (B) 462 (C) 151200 (D) 332640

B-9. A box contains 2 white balls, 3 black balls & 4 red balls. In how many ways can three balls be drawn fromthe box if atleast one black ball is to be included in draw (the balls of the same colour are different).(A) 60 (B) 64 (C) 56 (D) none

B-10. Passengers are to travel by a double decked bus which can accommodate 13 in the upper deck and 7 inthe lower deck. The number of ways that they can be distributed if 5 refuse to sit in the upper deck and 8refuse to sit in the lower deck is(A) 25 (B) 21 (C) 18 (D) 15

B-11*. In an examination, a candidate is required to pass in all the four subjects he is studying. The number ofways in which he can fail is(A) 4P1 + 4P2 + 4P3 + 4P4 (B) 44 – 1(C) 24 – 1 (D) 4C1 + 4C2 + 4C3 + 4C4

B-12*. In a cricket match against Zimbabwe, Azhar wants to bat before Jadeja and Jadeja wants to bat beforeGanguli. Number of possible batting orders with the above restrictions, if the remaining eight team membersare prepared to bat at any given place, is:

(A) 113

!! (B) 11C3. 8 ! (C)

113

!(D) none

B-13*. The number of ways of arranging the letters AAAAA, BBB, CCC, D, EE & F in a row if the letter C areseparated from one another is:

(A) 13C3. !2!3!5!12

(B) !2!3!3!5!13

(C) !2!3!3!14

(D) 11. !6!13

B-14*. 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 - 24C4 (B) 24C5 (C) 25C5 - 24C5 (D) 24C4

Section (C) : Problem base on distinct and identical objects/devisors (PCTS/PCDV)

C-1. The number of divisors of apbqcrds where a, b, c, d are primes & p, q, r, s Î N, excluding 1 and the numberitself is:(A) p q r s (B) (p + 1) (q + 1) (r + 1) (s + 1) - 4(C) p q r s

- 2 (D) (p + 1) (q + 1) (r + 1) (s + 1) - 2C-2. N is a lease natural number having 24 divisors. Then the number of ways N can be resolved into two factors

is(A) 12 (B) 24 (C) 6 (D) None of these

C-3. How many divisors of 21600 are divisible by 10 but not by 15?(A) 10 (B) 30 (C) 40 (D) none

C-4. The sum of the divisors of 25 . 37 . 53 . 72 is(A) 26 . 38 . 54 . 73 (B) 26 . 38 . 54 . 73 – 2 . 3 . 5 . 7(C) 26 . 38 . 54 . 73 – 1 (D) none of these

C-5. The number of ways in which the number 27720 can be split into two factors which are co-primes is:(A) 15 (B) 16 (C) 25 (D) 49


Section (D) : Problem based on circular arrangement/Multinomial theorem (PCMT/PCCA)

D-1. The number of ways in which 8 different flowers can be strung to form a garland so that 4 particularsflowers are never separated is:

(A) 4 !. 4 ! (B) 84

!!

(C) 288 (D) none

D-2. The number of ways in which 6 red roses and 3 white roses (all roses different) can form a garland so thatall the white roses come together is(A) 2170 (B) 2165 (C) 2160 (D) 2155

D-3. The number of ways in which 4 boys & 4 girls can stand in a circle so that each boy and each girl is oneafter the other is:(A) 3 !. 4 ! (B) 4 !. 4 ! (C) 8 ! (D) 7 !

D-4. The number of ways in which 5 beads, chosen from 8 different beads be threaded on to a ring is:(A) 672 (B) 1344 (C) 336 (D) none

D-5. Number of ways in which 3 persons throw a normal die to have a total score of 11 is(A) 27 (B) 25 (C) 29 (D) 18

D-6. If chocolates of a particular brand are all identical then the number of ways in which we can choose6 chocolates out of 8 different brands available in the market is:.(A) 13C6 (B) 13C8 (C) 86 (D) none

D-7. Number of positive integral solutions of x1 . x2 . x3 = 30 is(A) 25 (B) 26 (C) 27 (D) 28

Section (E) : Problem based on geometry/Dearrangement/exponent of prime/Principal of exculusion/Grouping (PCGT/PCDA/PCGP)

E-1. Number of ways in which 9 different toys be distributed among 4 children belonging to different age groupsin such a way that distribution among the 3 elder children is even and the youngest one is to receive onetoy more, is:

(A) ( )58

2!(B)

92!

(C) ( )9

3 2 3!

! !(D) none

E-2* 50C36 is divisible by(A) 19 (B) 52 (C) 192 (D) 53

E-3*. 2nPn is equal to(A) (n + 1) ( n + 2) ..... (2n) (B) 2n [1 . 3 . 5 .....(2n – 1)](C) (2) . (6) . (10) .... (4n – 2) (D) n! (2nCn)

E-4*. There are 12 points in a plane of which 5 are collinear. The number of distinct quadrilaterals which can beformed with vertices at these points is:(A) 2. 7P3 (B) 7P3 (C) 10. 7C3 (D) 420

E-5*. The number of ways in which 200 different things can be divided into groups of 100 pairs is:

(A)2002100

!(B)

1012

æèç

öø÷

1022

æèç

öø÷

1032

æèç

öø÷ ....

2002

æèç

öø÷

(C)200

2 100100!

( ) !(D) (1. 3. 5...... 199)


PART - II : SUBJECTIVE QUESTIONS

Section (A) : Problem based, Arrangements of given objects/Selection of given object (PCAD/PCSD)A-1. How many 3 digit even number can be formed using the digits 1, 2, 3, 4, 5 (repetition allowed).A-2. There are 10 buses operating between places A and B. In how many ways a person can go from place A to

place B and return to place A, if he returns in a different bus.A-3. The digits from 0 to 9 are written on slips of paper and placed in a box. Four of the slips are drawn at

random and placed in the order. How many out comes are possible.A-4. Find the number of 6 digit numbers that ends with 21 (eg. 537621), without repetition of digits.A-5. When a coin is tossed n times find the number of possible out comes.

A-6. (i) If !91

+ !101

= !11x

, find 'x'.

(ii) If nP5 = 42. nP3, find 'n'.A-7. How many words can be formed by using all the letters of the word 'MONDAY' if each word start with a

consonant.A-8. Find the number of natural numbers 1 to 1000 having none of their digits repeated.A-9. (i) If nC3 = nC5, find the value of nC2.

(ii) If 2nC3 : nC3 = 11 : 1, find 'n'.

(iii) If n – 1Cr : nCr :

n + 1Cr = 6 : 9 : 13, find n & r.

Section (B) : Problem based on Selection as well as arrangement of objects/Rank of word (PCSR/PCSI)

B-1. Find the number of words those can be formed by using all letters of the word 'DAUGHTER', if all thevowels must not be together.

B-2. A number lock has 4 dials, each dial has the digits 0, 1, 2, ........, 9. What is the maximum unsuccessfulattempts to open the lock.

B-3. If all the letters of the word 'AGAIN' are arranged in all possible ways & put in dictionary order, what is the50th word.

B-4. How many different permutations are possible using all the letters of the word MISSISSIPPI, if no two I'sare together.

B-5. Six persons meet in a room and each shakes hands with all the others. Find the total number of handshakes that took place.

B-6. In how many ways we can select a committee of 6 persons from 6 boys and 3 girls, if atleast two boys &atleast two girls must be there in the committee.

B-7. In how many ways playing 11 can be selected from 15 players, if only 6 of these players can bowl and theplaying 11 must include atleast 4 bowlers.

B-8. In a question paper there are two parts part A and part B each consisting of 5 questions. In how manyways a student can answer 6 questions, by selecting atleast two from each part.

B-9. How many four digit natural numbers not exceeding the number 4321 can be formed using the digits 1, 2,3, 4, if repetition is allowed.

B-10. A committee of 6 is to be chosen from 10 persons with the condition that if a particular person 'A' ischosen, then another particular person B must be chosen.

B-11. In how many ways a team of 5 can be chosen from 4 girls & 7 boys, if the team has atleast 3 girls.B-12. In how many ways we can select 5 cards from a deck of 52 cards, if each selection must include atleast

one king.B-13. Find ways of selection of atleast one vowel and one consonent from the word TRIPLE


Section (C) : Problem base on distinct and identical objects/devisors (PCTS/PCDV)

C-1. Let N = 24500, then find(i) The number of ways by which N can be resolved into two factors.(ii) The number of ways by which 5N can be resolved into two factors.(iii) The number of ways by which N can be resolved into two coprime factors.

C-2. Find the number of ways in which one or more letter be selected from the letters AAAABBCCCDEF.C-3. Find number of divisiors of 1980.

(i) How many of them are multiple of 11 ? find their sum .(ii) How many of them are divisible by 4 but not by 15.

Section (D) : Problem based on circular arrangement/Multinomial theorem (PCMT/PCCA)

D-1. There are 3 white, 4 blue and 1 red flowers. All of them are taken out one by one and arranged in a row inthe order. How many different arrangements are possible (flowers of same colurs are similar).

D-2. In how many ways 5 persons can sit at a round table, if two of the persons does not sit together.D-3. In how many ways four men and three women may sit around a round table if all the women are togetherD-4. Seven persons including A, B, C are seated on a circular table. How many arrangements are possible if B

is always between A and C.D-5. In how many ways four '+' and five '–' sign can be arranged in a circles so that no two '+' sign are together.D-6. How many ways fifteen different item may be given to A, B, C such that A gets 3, B gets 5 and remaining

goes to C.D-7. Find ways of distrubuting 8 different items equally among two children.D-8. Find number of negative integral solution of equation x + y + z = – 12D-9. In how many ways it is possible to divide six identical green, six identical blue and six identical red among

two persons such that each gets equal number of item.D-10. Find total number of positive integral solutions of 15 < x1 + x2 + x3 £ 20.

Section (E) : Problem based on geometry/Dearrangement/exponent of prime/Principal of exculusion/Grouping (PCGT/PCDA/PCGP)

E-1. In how many ways 18 diffrent objects can be divided into 7 groups such that four groups contains 3 objects eachand three groups contains 2 objects each.

E-2. (a) In how many ways can five people be divided into three groups?(b) In how many ways can five people be distributed in three different rooms if no room must be empty?(c) In how many ways can five people be arranged in three different rooms if no room must be empty?

E-3. Prove that : 20

200!(10!) 19! is an integer

E-4. Find exponent of 3 in 20 !E-5. Find number of zeros at the end of 45!.E-6. A person writes letters to five friends and addresses the corresponding envelops. In how many ways can

the letters be placed in the envelops so that(a) all letters are in the wrong envelops(b) at least three of them are in the wrong envelops.


PART - III : MISCELLANEOUS OBJECTIVE QUESTIONS

1. Consider the word "HONOLULU".Column – I Column – II(A) Number of words that can be formed using (p) 26

the letters of the given word in which consonants& vowels are alternate is

(B) Number of words that can be formed without (q) 144changing the order of vowels is

(C) Number of ways in which 4 letters can be (r) 840selected from the letters of the given word is

(D) Number of words in which two O's are together (s) 900but U's are separated is

2. Column – I Column – II

(A) The total number of selections of fruits which can be made (p) Greater than 50from, 3 bananas, 4 apples and 2 oranges is

(B) If 7 points out of 12 are in the same straight line, then (q) Greater than 100the number of triangles formed is

(C) The number of ways of selecting 10 balls from unlimited (r) Greater than 150number of red, black, white and green balls is

(D) The total number of proper divisors of 38808 is (s) Greater than 200

(COMPREHENSSION)

Comprehenssion - 1There are 8 official 4 non-official members, out of these 12 members a committee of 5 members is to be fromed,then answer the following questions.

3. Number of committees consisting of 3 official and 2 non-official members, are(A) 363 (B) 336 (C) 236 (D) 326

4. Number of committees consisting of at least two non-official members, are(A) 456 (B) 546 (C) 654 (D) 466

5. Number of committees in which a particular official member is never included, are(A) 264 (B) 642 (C) 266 (D) 462

Comprehenssion - 2Let n be the number of ways in which the letters of the word "RESONANCE" can be arranged so that vowelsappear at the even places and m be the number of ways in which "RESONANCE" can be arrange so that lettersR, S, O, A, appear in the order same as in the word RESONANCE, then answer the following questions.

6. The value of n is(A) 360 (B) 720 (C) 240 (D) 840

7. The value of m is(A) 3780 (B) 3870 (C) 3670 (D) 3760

8. The exponent of 5 in n is(A) 88 (B) 178 (C) 358 (D) None of these


(ASSERTION/REASON)

9. STATEMENT-1 : )!1n()!1n(

-+ is divisible by 6 for some n Î N.

STATEMENT-2 : Product of three consecutive integers is divisible by 3!.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

10. Statement -1 : The maximum number of points of intersection of 8 circles is 56.Statement -2 : The maximum number of points into which 4 circles and 4 straight lines interseet, is 50.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

11. Statement -1 : If there are six letters L1, L2 , L3, L4, L5, L6 and their corresponding six envelopes E1, E2, E3,E4, E5, E6. Letters having odd value can be put into odd value envelopes and even valueletters can be put into even value envelopes, so that no letter go into the right envelopes, thenumber of arrangement will be equal to 4.

Statement -2 : If Pn number of ways in which n letter can be put in ‘n’ corresponding envelopes such that no

letters goes to correct envelopes then Pn = n! ÷÷ø

öççè

æ+++-

!n)1(....

!21

!111

n


12. Statement -1 : The maximum value of K such that (50)k divides 100! is 2.

Statement -2 : If P is any prime number, then power of P in n! is equal to úûù

êëéPn

+ úûù

êëé

2Pn

+ úûù

êëé

3Pn

....

where [ · ] represents greatest integer function.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True

13. Statement-1: If a,b, c are positive integers such that a + b + c £ 8, then number of possible values of the orderedtriplets (a, b, c) is 56Statement-2: The number of ways in which n identical things can be distributed into r different groups isn–1 Cr–1


14. STATEMENT -1 : If N is number of positive integral solutions of x1 x2 x3 x4 = 770, then N is divisible by 4 distinct primes.

STATEMENT -2 : Prime numbers are 2, 3, 5, 7, 11, 13, .....(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement-1 is True, Statement-2 is False(D) Statement-1 is False, Statement-2 is True


PART - I : OBJECTIVE QUESTIONS

1. A train is going from London to Cambridge stops at 12 intermediate stations. 75 persons enter the trainduring the journey with 75 different tickets of the same class. Number of different sets of tickets they maybe holding is:(A) 78C3 (B) 91C75 (C) 84C75 (D) none

2. In 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

3. 12 guests at a dinner party are to be seated along a circular table. Supposing that the master andmistress of the house have fixed seats opposite one another, and that there are two specified guests whomust always, be placed next to one another; the number of ways in which the company can be placed, is:(A) 20. 10 ! (B) 22. 10 ! (C) 44. 10 ! (D) none

4. Six persons A, B, C, D, E and F are to be seated at a circular table. The number of ways this can be doneif 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

5. Out of 16 players of a cricket team, 4 are bowlers and 2 are wicket keepers. A team of 11 players is to bechosen so as to contain at least 3 bowlers and at least 1 wicketkeeper. The number of ways in which theteam be selected is(A) 2400 (B) 2472 (C) 2500 (D) 960

6. The number of ways in which 15 apples & 10 oranges can be distributed among three persons, eachreceiving none, one or more is:(A) 5670 (B) 7200 (C) 8976 (D) none of these

7. The number of permutations which can be formed out of the letters of the word "SERIES" taking threeletters together is:(A) 120 (B) 60 (C) 42 (D) none

8. Seven different coins are to be divided amongst three persons. If no two of the persons receive the samenumber of coins but each receives atleast one coin & none is left over, then the number of ways in whichthe division may be made is:(A) 420 (B) 630 (C) 710 (D) none

9. The streets of a city are arranged like the lines of a chess board. There are m streets running North toSouth & 'n' streets running East to West. The number of ways in which a man can travel from NW to SEcorner going the shortest possible distance is:

(A) m n2 2+ (B) ( ) . ( )m n- -1 12 2 (C) ( ) !

! . !m nm n

+(D)

( ) !( ) ! . ( ) !

m nm n

+ -- -

21 1

10. The number of ways in which a mixed double tennis game can be arranged from amongst 9 married coupleif no husband & wife plays in the same game is:(A) 756 (B) 3024 (C) 1512 (D) 6048

11. The number of ways in which 5 X's can be placed in the squares of the figure so that no row remains emptyis:

(A) 97 (B) 44 (C) 100 (D) 126


12. In a conference 10 speakers are present. If S1 wants to speak before S2 & S2 wants to speak afterS3, 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

!

13. If all the letters of the word "QUEUE" are arranged in all possible manner as they are in a dictionary, thenthe rank of the word QUEUE is:(A) 15th (B) 16th (C) 17th (D) 18th

14. Two variants of a test paper are distributed among 12 students. Number of ways of seating of the studentsin two rows so that the students sitting side by side do not have identical papers & those sitting in thesame column have the same paper is:

(A) 12

6 6!

! !(B)

( )!. !

122 65 (C) (6 !)2. 2 (D) 12 ! × 2

15. Sum of all the numbers that can be formed using all the digits 2, 3, 3, 4, 4, 4 is:(A) 22222200 (B) 11111100 (C) 55555500 (D) 20333280

16. Six married couple are sitting in a room. Number of ways in which 4 people can be selected so that thereis exactly one married couple among the four is:(A) 240 (B) 255 (C) 360 (D) 480

17. The number of ways selecting 8 books from a library which has 10 books each of Mathematics, Physics,Chemistry and English, if books of the same subject are alike, is:(A) 13C4 (B) 13C3 (C) 11C4 (D) 11C3

18. The number of integers which lie between 1 and 106 and which have the sum of the digits equal to 12 is:(A) 8550 (B) 5382 (C) 6062 (D) 8055

19. In a shooting competition a man can score 0, 2 or 4 points for each shot. Then the number of differentways in which he can score 14 points in 5 shots, is:(A) 20 (B) 24 (C) 30 (D) none

20. Number of ways in which a pack of 52 playing cards be distributed equally among four players so thateach may have the Ace, King, Queen and Jack of the same suit is:

(A) ( )369 4

!!

(B) ( )

36 49 4! . !!

(C) ( )

369 44

!! . !

(D) none

21. A box contains 6 balls which may be all of different colours or three each of two colours or two each ofthree different colours. The number of ways of selecting 3 balls from the box (if ball of same colour areidentical), is:(A) 60 (B) 31 (C) 30 (D) none

22. Number of ways in which 2 Indians, 3 Americans, 3 Italians and 4 Frenchmen can be seated on a circle,if the people of the same nationality sit together is:(A) 2. (4 !)2 (3 !)2 (B) 2. (3 !)3. 4 ! (C) 2. (3 !) (4 !)3 (D) none

23. Number of ways in which all the letters of the word " ALASKA " can be arranged in a circle distinguishingbetween the clockwise and anticlockwise arrangement, is:(A) 60 (B) 40 (C) 20 (D) none of these

24. Let Pn denotes the number of ways of selecting 3 people out of ' n ' sitting in a row, if no two of them areconsecutive and Qn is the corresponding figure when they are in a circle. If Pn - Qn = 6, then ' n ' is equal to:(A) 8 (B) 9 (C) 10 (D) 12

Multiple choice25. The number of non-negative integral solutions of x1 + x2 + x3 + x4 £ n (where n is a positive integer) is

(A) n+3C3 (B) n+4C4 (C) n+5C5 (D) n+4Cn


26. 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) 13C10 – 5C3 (D) 5C3 .

8C7 + 5C4 . 8C6 + 8C5

27. You are given 8 balls of different colour (black, white,...). The number of ways in which these balls can bearranged in a row so that the two balls of particular colour (say red & white) may never come together is:(A) 8 ! - 2.7 ! (B) 6. 7 ! (C) 2. 6 ! . 7C2 (D) none

28. There are 10 seats in the first row of a theatre of which 4 are to be occupied. The number of ways ofarranging 4 persons so that no two persons sit side by side is:(A) 7C4 (B) 4. 7P3 (C) 7C3. 4 ! (D) 840

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

(A)n -æ

èçöø÷

12

2

if n is even (B)( )n n - 2

4 if n is odd (C)

( )n -14

2

if n is odd (D)( )n n - 2

4 if n is even

30. The number of ways in which 10 students can be divided into three teams, one containing 4 and others 3each, is

(A) !3!3!4

!10(B) 2100 (C) 10C4 .

5C3 (D) !3!3!6

!10.

21

PART - II : SUBJECTIVE QUESTIONS

1. A family consists of a grandfather, m sons and daughters and 2n grand children. They are to be seated ina row for dinner. The grand children wish to occupy the n seats at each end and the grandfather refuses tohave a grand children on either side of him. In how many ways can the family be made to sit?

2. How many five digits numbers divisible by 3 can be formed using the digits 0, 1, 2, 3, 4, 7 and 8 if, eachdigit is to be used atmost one.

3. In how many other ways can the letters of the word MULTIPLE be arranged ; (i) without changing the orderof the vowels (ii) keeping the position of each vowel fixed (iii) without changing the relative order/position ofvowels & consonants.

4. A bouquet from 11 different flowers is to be made so that it contains not less than three flowers. Findnumber of different ways of selecting flowers to form the bouquet.

5. If a = x1 x2 x3 and b = y1 y2y3 be two three digit numbers, then find number of pairs of a and b that can beformed so that a can be subtracted from b without borrowing.

6. Show that the number of combinations of n letters together out of 3n letters of which n are a and n are band the rest unlike is, (n + 2). 2n - 1.

7. The integer from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number ismarked (that is 1, 16, 31, .... etc.). This process in continued untill a number is reached which has alreadybeen marked, then find number of unmarked numbers.

8. Find the number of positive integral solutions of, (i) x2 - y2 = 352706 (ii) xyz = 216009. The sides AB, BC & CA of a triangle ABC have 3, 4 & 5 interior points respectively on them. Find the

number of triangles that can be constructed using these interior points as vertices.10. How many positive integers of n digits exist such that each digit is 1, 2, or 3? How many of these contain

all three of the digits 1, 2 and 3 atleast once ?11. Find the number of ways in which 8 non-identical apples can be distributed among 3 boys such that every

boy should get atleast 1 apple & atmost 4 apples.12. There are ' n ' straight line in a plane, no two of which are parallel and no three pass through the same point.

Their points of intersection are joined. Show that the number of fresh lines thus introduced is,

81 n (n - 1) (n - 2) (n - 3).


PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)

1. How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits sothat the odd digits occupy even positions? [IIT - 2000, 1](A) 16 (B) 36 (C) 60 (D) 180

2. Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. IfTn+1 – Tn = 21, then n equals [IIT – 2001](A) 5 (B) 7 (C) 6 (D) 4

3. Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is [IIT – 2001, 1](A) 14 (B) 16 (C) 12 (D) 8

4. The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacentlyis [IIT – 2002](A) 40 (B) 60 (C) 80 (D) 100

5. Prove by permutation or otherwise ( )( )n

2

!n!n

is an integer (n Î I+). [IIT – 2004, 2]

6. A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit length by drawing parallel lines as shownin the diagram, then the number of rectangles possible with odd side lengths is [IIT - 2005]

(A) (m + n – 1)2 (B) 4m+n–1 (C) m2 n2 (D) m(m + 1)n(n + 1)

7. If total number of runs scored in n matches is ÷øö

çèæ +

41n

(2n+1 – n – 2) where n > 1, and the runs scored in the

kth match are given by k. 2n+1–k, where 1 £ k £ n, find n [IIT – 2005 , 2]8. If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2 t4s2, then the number

of ordered pair (p, q) is [IIT – 2006, (3, –1)](A) 252 (B) 254 (C) 225 (D) 224

9. The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order asin an English dictionary. Then number of words that appear before the word COCHIN is [IIT - 2007](A) 360 (B) 192 (C) 96 (D) 48 [IIT - 2007]

10. Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by dark-ening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I Column II(A) The number of permutations containing the word ENDEA is (p) 5!(B) The number of permutations in which the letter E occurs in the first (q) 2 × 5!

and the last positions is(C) The number of permutations in which none of the letters D, L, N occurs (r) 7 × 5!

in the last five positions is(D) The number of permutations in which the letters A, E, O occur only (s) 21 × 5!

in odd positions is


11. Then number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and3 only, is : [IIT-JEE- 2009,(3, -1) out of 82](A) 55 (B) 66 (C) 77 (D) 88

12. Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to :[IIT-JEE 2010, (5, -2) Out of 84]

(A) 25 (B) 34 (C) 42 (D) 4113. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that

each person gets at least one ball is : [IIT-JEE 2012, (3, –1) Out of 70](A) 75 (B) 150 (C) 210 (D) 243

Paragraph for Question No 14-15Let an denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutivedigits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and cn = the number of suchn-digit integers ending with digit 0. [IIT-JEE 2012, (3, –1) × 2 Out of 66]

14. Which of the following is correct ?(A) a17 = a16 + a15 (B) c17 ¹ c16 + c15 (C) b17 ¹ b16 + c16 (D) a17 = c17 + b16

15. The value of b6 is(A) 7 (B) 8 (C) 9 (D) 11

16. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack andthe sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards isk, then k – 20 = [JEE-Advanced 2013]

PART-II AIEEE (PREVIOUS YE ARS PROBLEMS)

1. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from thefirst five questions. The number of choices available to him is : [AIEEE 2003](1) 140 (2) 196 (3) 280 (4) 346

2. The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together,is given by : [AIEEE 2003](1) 6! × 5! (2) 30 (3) 5! × 4! (4) 7! × 5!

3. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order ?(1) 120 (2) 240 (3) 360 (4) 480 [AIEEE 2004]

4. The number of ways of distributing 8 identical balls in 3 distinct boxes, so that none of the boxes is empty, is:(1) 5 (2) 21 (3) 38 (4) 8C3 [AIEEE 2004]

5. If the letters of the word SACHIN are arranged in all possible ways and these words are written out as indictionary, then the word SACHIN appears at serial number : [AIEEE 2005](1) 602 (2) 603 (3) 600 (4) 601

6. The set S : = {1, 2, 3 ..........12} is to be partitioned into three sets A, B, C of equal size.Thus, A È B È C = S,

A Ç B = B Ç C = A Ç C = fThe number of ways to partition S is : [AIEEE 2007](1) 12!/3! (4!)3 (2) 12!/3!(3!)4 (3) 12!/(4!)3 (4) 12!/(3!)4

7. In a shop there are five types of ice-creams available.. A child buys six ice-creams. [AIEEE 2008]Statement-1 : The number of different ways the child can buy the six ice-creams, is 10C5.Statement-2 : The number of different ways the child can buy the six ice-creams is equal to the number of

different ways of arranging 6A' s and 4 B' s in a row.(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(3) Statement-1 is True, Statement-2 is False.(4) Statement-1 is False, Statement-2 is True.


8. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S areadjacent ? [AIEEE 2008](1) 8. 6C4 .

7C4 (2) 6. 7 8C4 (3) 6. 8. 7C4 (4) 7. 6C4 . 8C4

9. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged ina row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is:

[AIEEE 2009](1) atleast 500 but less than 750. (2) atleast 750 but less than 10000(3) atleast 1000. (4) less than 500.

10. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls aretaken out at random and then transferred to the other. The number of ways in which this can be done is :

[AIEEE 2010](1) 36 (2) 66 (3) 108 (4) 3

11. Statement-1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that nobox is empty is 9C3. [AIEEE 2011]

Statement-2 : The number of ways of choosing any 3 places from 9 different places is 9C3.(1) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.(2) Statement-1 is true, Statement-2 is false.(3) Statement-1 is false, Statement-2 is true.(4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

12. Assuming the balls to be identical except for difference in colours, the number of ways in which one or moreballs can be selected from 10 white, 9 green and 7 black balls is : [AIEEE 2012](1) 880 (2) 629 (3) 630 (4) 879

13. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets ofA × B having 3 or more elements is : [JEE-Mains 2013](1) 256 (2) 220 (3) 219 (4) 211

NCERT BOARD QUESTIONS

1. Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the womenchoose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the totalnumber of possible arrangements.

2. If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rankof the word RACHIT ?

3. A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, eachcontaining 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the numberof different ways of doing questions.

4. Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the numberof lines that can be formed joining the point.

5. We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many wayscan selections be made?

6. How many committee of five persons with a chairperson can be selected from 12 persons.

7. How many automobile license plates can be made if each plate contains two different letters followed by threedifferent digits?


8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can beselected from the lot.

9. Find the number of permutations of n distinct things taken r together, in which 3 particular things must occurtogether.

10. Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowelsare together.

11. Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided thatno digit is to be repeated.

12. There are 10 persons named P1, P2 ,P3, P4 .......... P10 Out of 10 persons, 5 persons are to be arranged in a linesuch that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possiblearrangements.

13. There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways inwhich the hall can be illuminated.

14. A box contains two white, three black and four red balls. In how many ways can three balls be drawn from thebox, if atleast one black ball is to be included in the draw.

15. If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.

16. Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digitsare repeated.

17. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how manypoints will they intersect each other?

18. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64.How many telephone numbers have all six digits distinct?

19. In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are howevercompulsory. Determine the number of ways in which the student can make the choice.

20. A convex polygon has 44 diagonals. Find the number of its sides.

Long Answer Type Questions

21. 18 mice were placed in two experimental groups and one control group, with all groups equally large. In howmany ways can the mice be placed into three groups?

22. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can bedrawn from the bag if (a) they can be of any colour (b) two must be white and two red and (c) they must all be ofthe same colour.

23. In how many ways can a football team of 11 players be selected from 16 players? How many of them will

(i) include 2 particular players?

(ii) exclude 2 particular players?

24. A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and atleast 5 from Class XII.If there are 20 students in each of these classes, in how many ways can the team be constituted?

25. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has

(i) no girls

(ii) at least one boy and one girl

(iii) at least three girls.


EXERCISE # 1

PART # I

A-1. (A) A-2. (C) A-3. (A) A-4. (B) B-1. (D) B-2. (D) B-3. (A)

B-4. (C) B-5. (B) B-6. (D) B-7. (C) B-8. (C) B-9. (B) B-10. (B)

B-11. (C, D) B-12. (A, B) B-13. (A, D) B-14. (A, B) C-1. (D) C-2. (A) C-3. (A)

C-4. (D) C-5. (B) D-1. (C) D-2. (C) D-3. (A) D-4. (A) D-5. (A)

D-6. (A) D-7. (C) E-1. (C) E-2* (A,B) E-3*. (A,B,C,D) E-4*. (A, D)

E-5*. (B,C,D)

PART # II

A-1. 50 A-2. 90 A-3. 10P4 A-4. 7. 7P3 A-5. 2n A-6. (i) 121 (ii) 10

A-7. 480 A-8. 738 A-9. (i) 28 (ii) 6 (iii) n = 12, r = 4.

B-1. 36000 B-2. 9999 B-3. NAAIG B-4. 7350 B-5. 15. B-6. 65 B-7. 1170

B-8. 200 B-9. 229 B-10. 154 B-11. 91 B-12. 886656 B-13. 45

C-1. (i) 18 (ii) 23 (iii) 4 C-2. 479

C-3. 36, (i) 18, 11.(20 + 21 + 22) (30 + 3 + 32) (5º + 5) (ii) 3.2 + 1.1.2 = 8

D-1. 280 D-2. 12 D-3. 144 D-4. 48 D-5. 1 D-6. 360360 D-7. 70

D-8. 55 D-9. 37 D-10. 685

E-1. 4 3

18!(3!) .(2i) 4!3! E-2. (a) 25 (b) 150 (c) 720 E-4. 8 E-5. 10

E-6. (a) = 44 (b) 109

PART # III1. (A) ® (q), (B) ® (r), (C) ® (p), (D) ® (s) 2. (A) ® (p), (B) ® (p, q, r), (C) ® (p, q, r, s), (D) ® (p)

3. (B) 4. (A) 5. (D) 6. (B) 7. (A) 8. (B) 9. (B)

10. (B) 11. (A) 12. (D) 13. (C) 14. (D)

EXERCISE # 2

PART # I

1. (A) 2. (A) 3. (A) 4. (D) 5. (B) 6. (C) 7. (C)

8. (B) 9. (D) 10. (C) 11. (B) 12. (D) 13. (C) 14. (D)

15. (A) 16. (A) 17. (D) 18. (C) 19. (C) 20. (B) 21. (B)

22. (B) 23. (C) 24. (C) 25. (B, D) 26. (A, C, D) 27. (A, B, C)

28. (B, C, D) 29. (C, D) 30. (B, C)


PART # II

1. (2n)! m! (m - 1) 2. 744 3. (i) 3359 (ii) 59 (iii) 359

4. 1981 5. 45.(55)2 7. 800 8. (i) Zero (ii) 1260 9. 205

10. 3n , 3n – 3.2n + 3 11. 4620

EXERCISE # 3

PART # I

1. (C) 2. (B) 3. (A) 4. (A) 6. (C) 7. 7 8. (C)

9. (C) 10. (A) ® (p), (B) ® (s), (C) ® (q), (D) ® (q) 11. (C) 12. (D) 13. (B)

14. (A) 15. (B) 16. k – 20 = 5

PART # II

1. (2) 2. (1) 3. (3) 4. (2) 5. (4) 6. (3) 7. (4)

8. (4) 9. (3) 10. (3) 11. (4) 12. (4) 13. (3)

EXERCISE # 4

NCERT BOARD QUESTIONS

1. 1440 2. 481 3. 780 4. 144 5. 22 6. 3960 7. 4,68000

8. 200 9. n – 3Cr – 3 (r – 2) ! 3 ! 10. 14400 11. 112 15. r = 3 16. 192

17. 190 18. 8400 19. 3 20. 11 21. 3

18!(6i)

22. (a) 11C4 (b) 6C2 × 5C2 (c) 6C4 + 5C4 23. (i) 14C9 (ii) 14C11

24. 2(20C5 × 20C6) 25. (i) 21, (ii) 441 (iii) 91

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