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Index1. Theory2. Short Revision3. Exercise (1 to 5)4. Assertion & Reason5. Que. from Compt. Exams
Subject : MathematicsTopic: Permutation &Combination
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2 of 2
7
Permutation and Combination
Permutations are arrangements and combinations are selections. In this chapter wediscuss the methods of counting of arrangements and selections. The basic resultsand formulas are as follows:
1. Fundamental Principle of Counting :
(i) Principle of Multiplication:If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’different ways, then total number of different ways of simultaneous occurrence of boththe events in a definite order is m × n.
(ii) Principle of Addition:If an event can occur in ‘m’ different ways, and another event can occur in ‘n’ differentways, then exactly one of the events can happen in m + n ways.
Example # 1There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur toDelhi. In how many ways a person can travel from Kota to Delhi via Jaipur by bus.
Solution.Let E1 be the event of travelling from Kota to Jaipur & E2 be the event of travelling fromJaipur to Delhi by the person.E1 can happen in 8 ways and E2 can happen in 10 ways.Since both the events E1 and E2 are to be happened in order, simultaneously, the numberof ways = 8 × 10 = 80.
Example # 2How many numbers between 10 and 10,000 can be formed by using the digits 1, 2, 3,4, 5 if(i) No digit is repeated in any number.(ii) Digits can be repeated.
Solution.(i) Number of two digit numbers = 5 × 4 = 20
Number of three digit numbers = 5 × 4 × 3 = 60Number of four digit numbers = 5 × 4 × 3 × 2 = 120
Total = 200
(ii) Number of two digit numbers = 5 × 5 = 25Number of three digit numbers = 5 × 5 × 5 = 125Number of four digit numbers = 5 × 5 × 5 × 5 = 625
Total = 775Self Practice Problems :1. How many 4 digit numbers are there, without repetition of digits, if each number is
divisible by 5.Ans. 952
2. Using 6 different flags, how many different signals can be made by using atleast three
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7flags, arranging one above the other.Ans. 1920
2. Arrangement :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 : (i) factorials of negative integers are not defined.(ii) 0 ! = 1 ! = 1 ;(iii) nPn = n ! = n. (n − 1) !(iv) (2n) ! = 2n. n ! [1. 3. 5. 7... (2n − 1)]
Example # 3How many numbers of three digits can be formed using the digits 1, 2, 3, 4, 5, withoutrepetition of digits. How many of these are even.
Solution.Three places are to be filled with 5 different objects.∴ Number of ways = 5P3 = 5 × 4 × 3 = 60For the 2nd part, unit digit can be filled in two ways & the remaining two digits can befilled in 4P2 ways.∴ Number of even numbers = 2 × 4P2 = 24.
Example # 4If all the letters of the word 'QUEST' are arranged in all possible ways and put in dictionaryorder, then find the rank of the given word.
Solution.Number of words beginning with E = 4P4 = 24Number of wards beginning with QE = 3P3 = 6Number of words beginning with QS = 6Number of words beginning withQT = 6.Next word is 'QUEST'∴ its rank is 24 + 6 + 6 + 6 + 1 = 43.
Self Practice Problems :3. Find the sum of all four digit numbers (without repetition of digits) formed using the
digits 1, 2, 3, 4, 5.Ans. 399960
4. Find 'n', if n – 1P3 : nP4 = 1 : 9.Ans. 9
5. Six horses take part in a race. In how many ways can these horses come in the first,second and third place, if a particular horse is among the three winners (Assume NoTies).Ans. 60
3. Circular Permutation :The number of circular permutations of n different things taken all at a time is; (n − 1) !.If clockwise & anti−clockwise circular permutations are considered to be same, then it
is2
)!1n( − .
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4 of 2
7Note: 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( −.
Example # 5In how many ways can we arrange 6 different flowers in a circle. In how many ways we can form agarland using these flowers.
Solution.The number of circular arrangements of 6 different flowers = (6 – 1)! = 120When we form a garland, clockwise and anticlockwise arrangements are similar. Therefore, the number
of ways of forming garland = 21
(6 – 1) ! = 60.
Example # 6In how many ways 6 persons can sit at a round table, if two of them prefer to sit together.
Solution.Let P1, P2, P3, P4, P5, P6 be the persons, where P1, P2 want to sit together.Regard these person as 5 objects. They can be arranged in a circle in (5 – 1)! = 24. Now P1P2 can bearranged in 2! ways. Thus the total number of ways = 24 × 2 = 48.
Self Practice Problems :
6. In how many ways the letters of the word 'MONDAY' can be written around a circle if the vowels are tobe separated in any arrangement.Ans. 72
7. In how many ways we can form a garland using 3 different red flowers, 5 different yellow flowers and 4different blue flowers, if flowers of same colour must be together.Ans. 17280.
4. Selection :
If nCr denotes the number of combinations of n different things taken r at a time, then
nCr = )!rn(!r!n− =
!rPr
n where r ≤ n ; n ∈ N and r ∈ W.
NOTE : (i) nCr = nCn – r(ii) nCr + nCr – 1 = n + 1Cr
(iii) nCr = 0 if r ∉ {0, 1, 2, 3........, n}
Example # 7Fifteen players are selected for a cricket match.(i) In how many ways the playing 11 can be selected(ii) In how many ways the playing 11 can be selected including a particular player.(iii) In how many ways the playing 11 can be selected excluding two particular players.
Solution.(i) 11 players are to be selected from 15
Number of ways = 15C11 = 1365.(ii) Since one player is already included, we have to select 10 from the remaining 14
Number of ways = 14C10 = 1001.(iii) Since two players are to be excluded, we have to select 11 from the remaining 13.
Number of ways = 13C11 = 78.Example # 8
If 49C3r – 2 = 49C2r + 1, find 'r'.Solution.
nCr = nCs if either r = s or r + s = n.Thus 3r – 2 = 2r + 1 ⇒ r = 3or 3r – 2 + 2r + 1 = 49 ⇒ 5r – 1 = 49 ⇒ r = 10∴ r = 3, 10
Example # 9A regular polygon has 20 sides. How many triangles can be drawn by using the vertices, but not usingthe sides.
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7Solution.The first vertex can be selected in 20 ways. The remaining two are to be selected from 17 vertices sothat they are not consecutive. This can be done in 17C2 – 16 ways.∴ The total number of ways = 20 × (17C2 – 16)But in this method, each selection is repeated thrice.
∴ Number of triangles = 3
)16C(20 217 −×
= 800.
Example # 1010 persons are sitting in a row. In how many ways we can select three of them if adjacent persons arenot selected.
Solution.Let P1, P2, P3, P4, P5, P6, P7, P8, P9, P10 be the persons sitting in this order.If three are selected (non consecutive) then 7 are left out.Let PPPPPPP be the left out & q, q, q be the selected. The number of ways in which these 3 q's canbe placed into the 8 positions between the P's (including extremes) is the number ways of requiredselection.Thus number of ways = 8C3 = 56.
Example # 11In how many ways we can select 4 letters from the letters of the word MΙSSΙSSΙPPΙ.
Solution.MΙΙΙΙSSSSPPNumber of ways of selecting 4 alike letters = 2C1 = 2.Number of ways of selecting 3 alike and 1 different letters = 2C1 × 3C1 = 6Number of ways of selecting 2 alike and 2 alike letters = 3C2 = 3Number of ways of selecting 2 alike & 2 different = 3C1 × 3C2 = 9Number of ways of selecting 4 different = 4C4 = 1Total = 21
Self Practice Problems :
8. In how many ways 7 persons can be selected from among 5 Indian, 4 British & 2 Chinese, if atleast twoare to be selected from each country.Ans. 100
9. 10 points lie in a plane, of which 4 points are collinear. Barring these 4 points no three of the 10 pointsare collinear. How many quadrilaterals can be drawn.Ans. 185.
10. In how many ways 5 boys & 5 girls can sit at a round table so that girls & boys sit alternate.Ans. 2880
11. In how many ways 4 persons can occupy 10 chairs in a row, if no two sit on adjacent chairs.Ans. 840.
12. In how many ways we can select 3 letters of the word PROPORTION.Ans. 36
5. 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 & of a third type & the remaining
n − (p + q + r) are all different is !r!q!p!n
.
Example # 12In how many ways we can arrange 3 red flowers, 4 yellow flowers and 5 white flowers in a row. In howmany ways this is possible if the white flowers are to be separated in any arrangement (Flowers of
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7same colour are identical).Solution.
Total we have 12 flowers 3 red, 4 yellow and 5 white.
Number of arrangements = !5!4!3!12
= 27720.
For the second part, first arrange 3 red & 4 yellow
This can be done in !4!3!7
= 35 ways
Now select 5 places from among 8 places (including extremes) & put the white flowers there.This can be done in 8C5 = 56.∴ The number of ways for the 2nd part = 35 × 56 = 1960.
Example # 13In how many ways the letters of the word "ARRANGE" can be arranged without altering the relativepositions of vowels & consonants.
Solution.
The consonants in their positions can be arranged in !2!4
= 12 ways.
The vowels in their positions can be arranged in !2!3 = 3 ways
∴ Total number of arrangements = 12 × 3 = 26
Self Practice Problems :
13. How many words can be formed using the letters of the word ASSESSMENT if each word begin with Aand end with T.Ans. 840
14. If all the letters of the word ARRANGE are arranged in all possible ways, in how many of words we willhave the A's not together and also the R's not together.Ans. 660
15. How many arrangements can be made by taking four letters of the word MISSISSIPPI.Ans. 176.
6. Formation of Groups :
Number of ways in which (m + n + p) different things can be divided into three different groups containing
m, n & p things respectively is( )
!p!n!m!pnm ++,
If m = n = p and the groups have identical qualitative characteristic 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(
.
Example # 1412 different toys are to be distributed to three children equally. In how many ways this can be done.
Solution.The problem is to divide 12 different things into three different groups.
Number of ways = !4!4!4!12
= 34650.
Example # 15
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7In how many ways 10 persons can be divided into 5 pairs.Solution.
We have each group having 2 persons and the qualitative characteristic are same (Since there is nopurpose mentioned or names for each pair).
Thus the number of ways = !5)!2(!10
5 = 945.
Self Practice Problems :
16. 9 persons enter a lift from ground floor of a building which stops in 10 floors (excluding ground floor). Ifis known that persons will leave the lift in groups of 2, 3, & 4 in different floors. In how many ways thiscan happen.Ans. 907200
17. In how many ways one can make four equal heaps using a pack of 52 playing cards.
Ans. !4)!13(!52
4
18. In how many ways 11 different books can be parcelled into four packets so that three of the packetscontain 3 books each and one of 2 books, if all packets have the same destination.
Ans. 2)!3(!114
7. Selection of one or more objects
(a) Number of ways in which atleast one object be selected out of 'n' distinct objects isnC1 + nC2 + nC3 +...............+ nCn = 2n – 1
(b) Number of ways in which atleast one object may be selected out of 'p' alike objects of one type'q' alike objects of second type and 'r' alike of third type is
(p + 1) (q + 1) (r + 1) – 1(c) Number of ways in which atleast one object may be selected from 'n' objects where 'p' alike of
one type 'q' alike of second type and 'r' alike of third type and restn – (p + q + r) are different, is(p + 1) (q + 1) (r + 1) 2n – (p + q + r) – 1
Example # 16There are 12 different books on a shelf. In how many ways we can select atleast one of them.
Solution.We may select 1 book, 2 books,........, 12 books.∴ The number of ways = 12C1 + 12C2 + ....... + 12C12 = 212 – 1. = 4095
Example # 17There are 12 fruits in a basket of which 5 are apples, 4 mangoes and 3 bananas (fruits of same speciesare identical). How many ways are there to select atleast one fruit.
Solution.Let x be the number of apples being selectedy be the number of mangoes being selected andz be the number of bananas being selected.Then x = 0, 1, 2, 3, 4, 5
y = 0, 1, 2, 3, 4z = 0, 1, 2, 3
Total number of triplets (x, y, z) is 6 × 5 × 4 = 120Exclude (0, 0, 0)∴ Number of combinations = 120 – 1 = 119.
Self Practice Problems
19. In a shelf there are 5 physics, 4 chemistry and 3 mathematics books. How many combinations arethere if (i) books of same subject are different (ii) books of same subject are identical.Ans. (i) 4095 (ii) 119
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720. From 5 apples, 4 mangoes & 3 bananas in how many ways we can select atleast two fruits of eachvariety if (i) fruits of same species are identical (ii) fruits of same species are different.Ans. (i) 24 (ii) 1144
8. Multinomial Theorem:Coefficient of xr in expansion of (1 − x)−n = n+r−1Cr (n ∈ N)Number of ways in which it is possible to make a selection from m + n + p = N things, where p are alikeof one kind, m alike of second kind & n alike of third kind taken r at a time is given by coefficient ofxr in the expansion of
(1 + x + x2 +...... + xp) (1 + x + x2 +...... + xm) (1 + x + x2 +...... + xn).
(i) For example the number of ways in which a selection of four letters can be made from theletters 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) (1 + x).
(ii) Method of fictious partition :Number of ways in which n identical things may be distributed among p persons if each personmay receive none, one or more things is; n+p−1Cn.
Example # 18Find the number of solutions of the equation x + y + z = 6, where x, y, z ∈ W.
Solution.Number of solutions = coefficient of x6 in (1 + x + x2 + ....... x6)3
= coefficient of x6 in (1 – x7)3 (1 – x)–3
= coefficient of x6 in (1 – x)–3
=
−+6
163 = 8C6 = 28.
Example # 19In a bakery four types of biscuits are available. In how many ways a person can buy 10 biscuits if hedecide to take atleast one biscuit of each variety.
Solution.Let x be the number of biscuits the person select from first variety, y from the second, z from the thirdand w from the fourth variety. Then the number of ways = number of solutions of the equationx + y + z + w = 10.where x = 1, 2, .........,7
y = 1, 2, .........,7z = 1, 2, .........,7w = 1, 2, .........,7
This is equal to = coefficient of x10 in (x + x2 + ...... + x7)4
= coefficient of x6 in (1 + x + ....... + x6)4
= coefficient of x6 in (1 – x7)4 (1 – x)–4
= coefficient x6 in (1 – x)–4
=
−+6
164 = 84.
Self Practice Problems:
21. Three distinguishable dice are rolled. In how many ways we can get a total 15.Ans. 10.
22. In how many ways we can give 5 apples, 4 mangoes and 3 oranges (fruits of same species are similar)to three persons if each may receive none, one or more.Ans. 3150
9. 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)........
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(c) Number of ways in which N can be resolved as a product of two factors is
= [ ] squareperfectaisNif1....)1c()1b()1a(squareperfectanotisNif....)1c()1b()1a(
21
21
+++++++
(d) Number of ways in which a composite number N can be resolved into two factors which arerelatively prime (or coprime) to each other is equal to 2n−1 where n is the number of differentprime factors in N.
Example # 20Find the number of divisors of 1350. Also find the sum of all divisors.
Solution.1350 = 2 × 33 × 52
∴ Number of divisors = (1+ 1) (3 + 1) (2 + 1) = 24sum of divisors = (1 + 2) (1 + 3 + 32 + 33) (1 + 5 + 52) = 3720.
Example # 21In how many ways 8100 can be resolved into product of two factors.
Solution.8100 = 22 × 34 × 52
Number of ways = 21
((2 + 1) (4 + 1) (2 + 1) + 1) = 23
Self Practice Problems :
23. How many divisors of 9000 are even but not divisible by 4. Also find the sum of all such divisors.Ans. 12, 4056.
24. In how many ways the number 8100 can be written as product of two coprime factors.Ans. 4
10. Let there be 'n' types of objects, with each type containing atleast r objects. Then the number of waysof arranging r objects in a row is nr.
Example # 22How many 3 digit numbers can be formed by using the digits 0, 1, 2, 3, 4, 5. In how many of these wehave atleast one digit repeated.
Solution.We have to fill three places using 6 objects (repeatation allowed), 0 cannot be at 100th place. The
number of numbers = 180.
Number of numbers in which no digit is repeated = 100∴ Number of numbers in which atleast one digit is repeated = 180 – 100 = 80
Example # 23How many functions can be defined from a set A containing 5 elements to a set B having 3 elements.How many these are surjective functions.
Solution.Image of each element of A can be taken in 3 ways.∴ Number of functions from A to B = 35 = 243.Number of into functions from A to B = 25 + 25 + 25 – 3 = 93.∴ Number of onto functions = 150.
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Self Practice Problems :
25. Find the sum of all three digit numbers those can be formed by using the digits. 0, 1, 2, 3, 4.Ans. 27200.
26. How many functions can be defined from a set A containing 4 elements to a set B containing 5 elements.How many of these are injective functions.Ans. 625, 120
27. In how many ways 5 persons can enter into a auditorium having 4 entries.Ans. 1024.
11. Dearrangement :Number of ways in which 'n' letters can be put in 'n' corresponding envelopes such that no letter goesto correct envelope is
n !
−++−+−
!n1)1(............
!41
!31
!21
!111 n
Example # 24In how many ways we can put 5 writings into 5 corresponding envelopes so that no writing go to thecorresponding envelope.
Solution.The problem is the number of dearragements of 5 digits.
This is equal to 5!
−+−
!51
!41
!31
!21
= 44.
Example # 25Four slip of papers with the numbers 1, 2, 3, 4 written on them are put in a box. They are drawn one byone (without replacement) at random. In how many ways it can happen that the ordinal number ofatleast one slip coincide with its own number.
Solution.Total number of ways = 4 ! = 24.The number of ways in which ordinal number of any slip does not coincide with its own number is the
number of dearrangements of 4 objects = 4 !
+−
!41
!31
!21
= 9
Thus the required number of ways. = 24 – 9 = 15
Self Practice Problems:
28. In a match column question, Column Ι contain 10 questions and Column II contain 10 answers writtenin some arbitrary order. In how many ways a student can answer this question so that exactly 6 of hismatchings are correct.Ans. 1890
29. In how many ways we can put 5 letters into 5 corresponding envelopes so that atleast one letter go towrong envelope.Ans. 119
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27ShorShorShorShorShort Rt Rt Rt Rt ReeeeevisionvisionvisionvisionvisionDEFINITIONS :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 different ways of simultaneous occurrence of both events in a definite order ism × 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. 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
nCr = )!rn(!r
!n−
=!rPr
n where r ≤ n ; n ∈ N and r ∈ 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!n
)!n2( ; for in any one way it is possible to interchange 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( ++ , m ≠ n ≠ p.
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(
.
(vi) The number of permutations of n things taken all at a time when p of them are similar & of one type, q ofthem are similar & of another type, r of them are similar & of a third type & the remaining
n – (p + q + r) are all different is : !r!q!p
!n .
(vii) The number of circular permutations of n different things taken all at a time is ; (n − 1)!. If clockwise &
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 ndistinct things.
(ix) Total number of ways in which it is possible to make a selection by taking some or all out ofp + 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
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of one kind , m alike of second kind & n alike of 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 ∈ N). For example the number of waysin which a selection of four letters can be made from the letters of the word PROPORTION is given bycoefficient of x4 in (1 + x + x2 + x3) (1 + x + x2) (1 + x + x2) (1 + x) (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 thenumber of things received by men = pn.
(xii) Number of ways in which n identical things may be distributed among p persons if each person mayreceive none , one or more things is ; n+p−1Cn.
(xiii) a. nCr = nCn−r ; nC0 = nCn = 1 ; b. nCx = nCy ⇒ x = y or x + y = n
c. nCr + nCr−1 = n+1Cr
(xiv) nCr is maximum if : (a) r = 2n if n is even. (b) r = 2
1n − or 21n + 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
factors is = [ ] squareperfectaisNif1)....1c)(1b)(1a(squareperfectanotisNif....)1c)(1b)(1a(
21
21
+++++++
(d) Number of ways in which a composite number N can be resolved into two factors which arerelatively prime (or coprime) to each other is equal to 2n−1 where n is the number of differentprime factors in N. [ Refer Q.No.28 of Ex−I ]
(xvi) Grid Problems and tree diagrams.
DEARRANGEMENT :Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own
envelope is = n!12!
13
14
1 1− + + + −
! !
........... ( )!
n
n .
(xvii) Some times students find it difficult to decide whether a problem is on permutation or combination orboth. Based on certain words / phrases occuring in the problem we can fairly decide its nature as per thefollowing table :PROBLEMS OF COMBINATIONS PROBLEMS OF PERMUTATIONS" Selections , choose " Arrangements" Distributed group is formed " Standing in a line seated in a row" Committee " problems on digits" Geometrical problems " Problems on letters from a word
EXEREXEREXEREXEREXERCISE–1CISE–1CISE–1CISE–1CISE–1Q.1 The straight lines l1 , l2 & l3 are parallel & lie in the same plane. A total of m points are taken on the line
l1 , n points on l2 & k points on l3. How many maximum number of triangles are there whose vertices areat these points ?
Q.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 once.
Q.3 There are 2 women participating in a chess tournament. Every participant played 2 games with the otherparticipants. The number of games that the men played between themselves exceeded by 66 as comparedto the number of games that the men played with the women. Find the number of participants & the totalnumbers of games played in the tournament.
Q.4 All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once, and not divisible by5 are arranged in the increasing order. Find the (2004)th number in this list.
Q.5 5 boys & 4 girls sit in a straight line. Find the number of ways in which they can be seated if 2 girls are
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together & the other 2 are also together but separate from the first 2.Q.6 A crew of an eight oar boat has to be chosen out of 11 men five of whom can row on stroke side only, four
on the bow side only, and the remaining two on either side. How many different selections can be made?Q.7 An examination paper consists of 12 questions divided into parts A & B.
Part-A contains 7 questions & Part−B contains 5 questions. A candidate is required to attempt 8 questionsselecting atleast 3 from each part. In how many maximum ways can the candidate select the questions ?
Q.8 In how many ways can a team of 6 horses be selected out of a stud of 16 , so that there shall always be3 out of A B C A ′ B ′ C ′ , but never A A ′ , B B ′ or C C ′ together.
Q.9 During a draw of lottery, tickets bearing numbers 1, 2, 3,......, 40, 6 tickets are drawn out & thenarranged in the descending order of their numbers. In how many ways, it is possible to have 4th ticketbearing number 25.
Q.10 Find the number of distinct natural numbers upto a maximum of 4 digits and divisible by 5, which can beformed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 each digit not occuring more than once in each number.
Q.11 The Indian cricket team with eleven players, the team manager, the physiotherapist and two umpires are totravel from the hotel where they are staying to the stadium where the test match is to be played. Four of themresiding in the same town own cars, each a four seater which they will drive themselves. The bus which was topick them up failed to arrive in time after leaving the opposite team at the stadium. In how many ways can theybe seated in the cars ? In how many ways can they travel by these cars so as to reach in time, if the seatingarrangement in each car is immaterial and all the cars reach the stadium by the same route.
Q.12 There are n straight lines in a plane, no 2 of which parallel , & no 3 pass through the same point. Theirpoint of intersection are joined. Show that the number of fresh lines thus introduced is
8)3n)(2n)(1n(n −−− .
Q.13 In how many ways can you divide a pack of 52 cards equally among 4 players. In how many ways thecards can be divided in 4 sets, 3 of them having 17 cards each & the 4th with 1 card.
Q.14 A firm of Chartered Accountants in Bombay has to send 10 clerks to 5 different companies, two clerksin each. Two of the companies are in Bombay and the others are outside. Two of the clerks prefer towork in Bombay while three others prefer to work outside. In how many ways can the assignment bemade if the preferences are to be satisfied.
Q.15 A train going from Cambridge to London stops at nine intermediate stations. 6 persons enter the trainduring the journey with 6 different tickets of the same class. How many different sets of ticket may theyhave had?
Q.16 Prove that if each of m points in one straight line be joined to each of n in another by straight linesterminated by the points, then excluding the given points, the lines will intersect
41 mn(m – 1)(n –1) times.
Q.17 How many arrangements each consisting of 2 vowels & 2 consonants can be made out of the letters ofthe word ‘DEVASTATION’?
Q.18 Find the number of words each consisting of 3 consonants & 3 vowels that can be formed from theletters of the word “Circumference”. In how many of these c’s will be together.
Q.19 There are 5 white , 4 yellow , 3 green , 2 blue & 1 red ball. The balls are all identical except for colour.These are to be arranged in a line in 5 places. Find the number of distinct arrangements.
Q.20 How many 4 digit numbers are there which contains not more than 2 different digits?Q.21 In how many ways 8 persons can be seated on a round table(a) If two of them (say A and B) must not sit in adjacent seats.(b) If 4 of the persons are men and 4 ladies and if no two men are to be in adjacent seats.(c) If 8 persons constitute 4 married couples and if no husband and wife, as well as no two men, are to be
in adjacent seats?Q.22 (i) If 'n' things are arranged in circular order , then show that the number of ways of selecting four of
the things no two of which are consecutive is
!4)7n()6n()5n(n −−−
(ii) If the 'n' things are arranged in a row, then show that the number of such sets of four is
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!4)6n()5n()4n()3n( −−−−
Q.23 (a) How many divisors are there of the number x = 21600. Find also the sum of these divisors.(b) In how many ways the number 7056 can be resolved as a product of 2 factors.(c) Find the number of ways in which the number 300300 can be split into 2 factors which are
relatively prime.Q.24 How many ten digits whole number satisfy the following property they have 2 and 5 as digits, and there
are no consecutive 2's in the number (i.e. any two 2's are separated by at least one 5).Q.25 How many different ways can 15 Candy bars be distributed between Ram, Shyam, Ghanshyam and
Balram, if Ram can not have more than 5 candy bars and Shyam must have at least two. Assume allCandy bars to be alike.
Q.26 Find the number of distinct throws which can be thrown with 'n' six faced normal dice which areindistinguishable among themselves.
Q.27 How many integers between 1000 and 9999 have exactly one pair of equal digit such as 4049 or 9902but not 4449 or 4040?
Q.28 In a certain town the streets are arranged like the lines of a chess board. There are 6 streets running north& south and 10 running east & west. Find the number of ways in which a man can go from the north-westcorner to the south-east corner covering the shortest possible distance in each case.
Q.29 (i) Prove that : nPr = n−1Pr + r. n−1Pr−1(ii) If 20Cr+2 = 20C2r−3 find 12Cr(iii) Find the ratio 20Cp to 25Cr when each of them has the greatest value possible.(iv) Prove that n−1C3 + n−1C4 > nC3 if n > 7.(v) Find r if 15C3r = 15Cr+3
Q.30 There are 20 books on Algebra & Calculus in our library. Prove that the greatest number of selectionseach of which consists of 5 books on each topic is possible only when there are 10 books on each topicin the library.
EXEREXEREXEREXEREXERCISE–2CISE–2CISE–2CISE–2CISE–2Q.1 Find the number of ways in which 3 distinct numbers can be selected from the set
{31, 32, 33, ....... 3100, 3101} so that they form a G.P.
Q.2 Let n & k be positive integers such that n ≥ 2
)1k(k + . Find the number of solutions
(x1 , x2 ,.... , xk) , x1 ≥ 1 , x2 ≥ 2 ,... , xk ≥ k , all integers, satisfying x1 + x2 + .... + xk = n.
Q.3 There are counters available in 7 different colours. Counters are all alike except for the colour and theyare atleast ten of each colour. Find the number of ways in which an arrangement of 10 counters can bemade. How many of these will have counters of each colour.
Q.4 For each positive integer k, let Sk denote the increasing arithmetic sequence of integers whose first termis 1 and whose common difference is k. For example, S3 is the sequence 1, 4, 7, 10...... Find the numberof values of k for which Sk contain the term 361.
Q.5 Find the number of 7 lettered words each consisting of 3 vowels and 4 consonants which can be formedusing the letters of the word "DIFFERENTIATION".
Q.6 A shop sells 6 different flavours of ice-cream. In how many ways can a customer choose 4 ice-creamcones if(i) they are all of different flavours(ii) they are non necessarily of different flavours(iii) they contain only 3 different flavours(iv) they contain only 2 or 3 different flavours?
Q.7 6 white & 6 black balls of the same size are distributed among 10 different urns. Balls are alike exceptfor the colour & each urn can hold any number of balls. Find the number of different distribution of theballs so that there is atleast 1 ball in each urn.
Q.8 There are 2n 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 not be placed next to one
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another. Show that the number of ways in which the company can be placed is (2n − 2)!.(4n2 − 6n + 4).Q.9 Each of 3 committees has 1 vacancy which is to be filled from a group of 6 people. Find the number of
ways the 3 vacancies can be filled if ;(i) Each person can serve on atmost 1 committee.(ii) There is no restriction on the number of committees on which a person can serve.(iii) Each person can serve on atmost 2 committees.
Q.10 How many 15 letter arrangements of 5 A's, 5 B's and 5 C's have no A's in the first 5 letters, no B's in thenext 5 letters, and no C's in the last 5 letters.
Q.11 5 balls are to be placed in 3 boxes. Each box can hold all 5 balls. In how many different ways can weplace the balls so that no box remains empty if,(i) balls & boxes are different (ii) balls are identical but boxes are different(iii) balls are different but boxes are identical (iv) balls as well as boxes are identical(v) balls as well as boxes are identical but boxes are kept in a row.
Q.12 In how many other ways can the letters of the word MULTIPLE be arranged;(i) without changing the order of the vowels(ii) keeping the position of each vowel fixed &(iii) without changing the relative order/position of vowels & consonants.Q.13 Find the number of ways in which the number 30 can be partitioned into three unequal parts, each part
being a natural number. What this number would be if equal parts are also included.Q.14 In an election for the managing committee of a reputed club , the number of candidates contesting
elections exceeds the number of members to be elected by r (r > 0). If a voter can vote in 967 differentways to elect the managing committee by voting atleast 1 of them & can vote in 55 different ways to elect(r − 1) candidates by voting in the same manner. Find the number of candidates contesting the elections& the number of candidates losing the elections.
Q.15 Find the number of three digits numbers from 100 to 999 inclusive which have any one digit that is theaverage of the other two.
Q.16 Prove by combinatorial argument that :(a) n+1Cr = nCr + nCr – 1(b) n + mcr = nc0 ·
mcr + nc1 · mcr − 1 + nc2 ·
mcr − 2 +....... + ncr · mc0.
Q.17 A man has 3 friends. In how many ways he can invite one friend everyday for dinner on 6 successivenights so that no friend is invited more than 3 times.
Q.18 12 persons are to be seated at a square table, three on each side. 2 persons wish to sit on the north sideand two wish to sit on the east side. One other person insists on occupying the middle seat (which maybe on any side). Find the number of ways they can be seated.
Q.19 There are 15 rowing clubs; two of the clubs have each 3 boats on the river; five others have each 2 andthe remaining eight have each 1; find the number of ways in which a list can be formed of the order of the24 boats, observing that the second boat of a club cannot be above the first and the third above thesecond. How many ways are there in which a boat of the club having single boat on the river is at thethird place in the list formed above?
Q.20 25 passengers arrive at a railway station & proceed to the neighbouring village. At the station there are2 coaches accommodating 4 each & 3 carts accommodating 3 each. Find the number of ways in whichthey can proceed to the village assuming that the conveyances are always fully occupied & that theconveyances are all distinguishable from each other.
Q.21 An 8 oared boat is to be manned by a crew chosen from 14 men of which 4 can only steer but can notrow & the rest can row but cannot steer. Of those who can row, 2 can row on the bow side. In howmany ways can the crew be arranged.
Q.22 How many 6 digits odd numbers greater than 60,0000 can be formed from the digits 5, 6, 7, 8, 9, 0 if(i) repetitions are not allowed (ii) repetitions are allowed.
Q.23 Find the sum of all numbers greater than 10000 formed by using the digits 0 , 1 , 2 , 4 , 5 no digit beingrepeated in any number.
Q.24 The members of a chess club took part in a round robin competition in which each plays every one elseonce. All members scored the same number of points, except four juniors whose total score were 17.5.How many members were there in the club? Assume that for each win a player scores 1 point , for draw1/2 point and zero for losing.
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27Q.25 There are 3 cars of different make available to transport 3 girsls and 5 boys on a field trip. Each car canhold up to 3 children. Find
(a) the number of ways in which they can be accomodated.(b) the numbers of ways in which they can be accomodated if 2 or 3 girls are assigned to one of the cars.
In both the cars internal arrangement of childrent inside the car is to be considered as immaterial.Q.26 Six faces of an ordinary cubical die marked with alphabets A, B, C, D, E and F is thrown n times and the
list of n alphabets showing up are noted. Find the total number of ways in which among the alphabetsA, B, C, D, E and F only three of them appear in the list.
Q.27 Find the number of integer betwen 1 and 10000 with at least one 8 and at least one 9 as digits.Q.28 The number of combinations n together of 3n letters of which n are 'a' and n are 'b' and the rest unlike is
(n + 2). 2n − 1.Q.29 In Indo−Pak one day International cricket match at Sharjah , India needs 14 runs to win just before the
start of the final over. Find the number of ways in which India just manages to win the match (i.e. scoresexactly 14 runs) , assuming that all the runs are made off the bat & the batsman can not score more than4 runs off any ball.
Q.30 A man goes in for an examination in which there are 4 papers with a maximum of m marks for eachpaper; show that the number of ways of getting 2m marks on the whole is
31 (m + 1) (2m² + 4m + 3).
EXEREXEREXEREXEREXERCISE–3CISE–3CISE–3CISE–3CISE–3Q.1 Find the total number of ways of selecting five letters from the letters of the word INDEPENDENT.
[ REE '97, 6 ]Q.2 Select the correct alternative(s). [ JEE ’98, 2 + 2 ](i) Number of divisors of the form 4n + 2 ( n ≥ 0) of the integer 240 is
(A) 4 (B) 8 (C) 10 (D) 3
(ii) An n-digit number is a positive number with exactly 'n' digits. Nine hundred distinct n-digit numbers areto be formed using only the three digits 2, 5 & 7. The smallest value of n for which this is possible is :(A) 6 (B) 7 (C) 8 (D) 9
Q.3 How many different nine digit numbers can be formed from the number 223355888 by rearranging itsdigits so that the odd digits occupy even positions ? [JEE '2000, (Scr)](A) 16 (B) 36 (C) 60 (D) 180
Q.4 Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of' n ' sides. If Tn + 1 − Tn = 21 , then ' n ' equals: [ JEE '2001, (Scr) ](A) 5 (B) 7 (C) 6 (D) 4
Q.5 The number of arrangements of the letters of the word BANANA in which the two N’s do not appearadjacently is [JEE 2002 (Screening), 3](A) 40 (B) 60 (C) 80 (D) 100
Q.6 Number of points with integral co-ordinates that lie inside a triangle whose co-ordinates are(0, 0), (0, 21) and (21,0) [JEE 2003 (Screening), 3](A) 210 (B) 190 (C) 220 (D) None
Q.7 Using permutation or otherwise, prove that n
2
)!n(!)n( is an integer, where n is a positive integer.
[JEE 2004, 2 out of 60]Q.8 A rectangle with sides 2m – 1 and 2n – 1 is divided into squares of unit lengthby drawing parallel lines as shown in the diagram, then the number of rectanglespossible with odd side lengths is
(A) (m + n + 1)2 (B) 4m + n – 1
(C) m2n2 (D) mn(m + 1)(n + 1)[JEE 2005 (Screening), 3]
Q.9 If r, s, t are prime numbers and p, q are the positive integers such that their LCM of p, q is is r2t4s2, thenthe numbers of ordered pair of (p, q) is(A) 252 (B) 254 (C) 225 (D) 224 [JEE 2006, 3]
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17 of
27EXEREXEREXEREXEREXERCISE–4CISE–4CISE–4CISE–4CISE–4Part : (A) Only one correct option1. There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number
of 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
2. 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.
3. The number of ways in which n different things can be given to r persons when there is no restriction asto the number of things each may receive is:(A) nCr (B) nPr (C) nr (D) rn
4. The number of divisors of apbqcrds where a, b, c, d are primes & p, q, r, s ∈ N, excluding 1 and thenumber itself 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) − 2
5. The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100is:(A) 3125 (B) 5081 (C) 6005 (D) 4851
6. Number 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 thesame side, is (k). 5 ! then k has the value equal to:(A) 2 (B) 4 (C) 8 (D) none
7. 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
8. 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
9. 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
10. 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
11. 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 waysin which the division may be made is:(A) 420 (B) 630 (C) 710 (D) none
12. 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
13. 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
!
14. Two variants of a test paper are distributed among 12 students. Number of ways of seating of thestudents in two rows so that the students sitting side by side do not have identical papers & thosesitting in the same column have the same paper is:
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27(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. There are m apples and n oranges to be placed in a line such that the two extreme fruits being bothoranges. Let P denotes the number of arrangements if the fruits of the same species are different andQ the corresponding figure when the fruits of the same species are alike, then the ratio P/Q has thevalue equal to:(A) nP2. mPm. (n − 2) ! (B) mP2. nPn. (n − 2) ! (C) nP2. nPn. (m − 2) ! (D) none
17. 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
18. 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)
( )36
9 44!
! . !(D) none
19. A five letter word is to be formed such that the letters appearing in the odd numbered positions aretaken from the letters which appear without repetition in the word "MATHEMATICS". Further the lettersappearing in the even numbered positions are taken from the letters which appear with repetition in thesame word "MATHEMATICS". The number of ways in which the five letter word can be formed is:(A) 720 (B) 540 (C) 360 (D) none
20. Number of ways of selecting 5 coins from coins three each of Rs. 1, Rs. 2 and Rs. 5 if coins of thesame denomination are alike, is:(A) 9 (B) 12 (C) 21 (D) none
21. 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
22. 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 thenumber of ordered pair (p, q) is [IIT – 2006](A) 252 (B) 254 (C) 225 (D) 224
Part : (B) May have more than one options correct
23. n + 1C6 + nC4 > n + 2C5 − nC5 for all ' n ' greater than:(A) 8 (B) 9 (C) 10 (D) 11
24. In an examination, a candidate is required to pass in all the four subjects he is studying. The numberof ways in which he can fail is(A) 4P1 + 4P2 + 4P3 + 4P4 (B) 44 – 1(C) 24 – 1 (D) 4C1 + 4C2 + 4C3 + 4C4
25. The kindergarten teacher has 25 kids in her class. She takes 5 of them at a time, to zoological gardenas often as she can, without taking the same 5 kids more than once. Then the number of visits, theteacher makes to the garden exceeds that of a kid by:(A) 25C5 − 24C4 (B) 24C5 (C) 25C5 − 24C5 (D) 24C4
26. 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
27. There are 10 points P1, P2,...., P10 in a plane, no three of which are collinear. Number of straight lineswhich can be determined by these points which do not pass through the points P1 or P2 is:(A) 10C2 − 2. 9C1 (B) 27 (C) 8C2 (D) 10C2 − 2. 9C1 + 1
28. Number of quadrilaterals which can be constructed by joining the vertices of a convex polygon of20 sides if none of the side of the polygon is also the side of the quadrilateral is:
(A) 17C4 − 15C2 (B) 15
3 204
C .(C) 2275 (D) 2125
29. You are given 8 balls of different colour (black, white,...). The number of ways in which these balls canbe arranged in a row so that the two balls of particular colour (say red & white) may never cometogether is:(A) 8 ! − 2.7 ! (B) 6. 7 ! (C) 2. 6 !. 7C2 (D) none
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2730. A man is dealt a poker hand (consisting of 5 cards) from an ordinary pack of 52 playing cards. Thenumber of ways in which he can be dealt a "straight" (a straight is five consecutive values not of thesame suit, eg. {Ace, 2, 3, 4, 5}, {2, 3, 4, 5, 6}.......................... & {10, J, Q, K, Ace}) is(A) 10 (45 − 4) (B) 4 !. 210 (C) 10. 210 (D) 10200
31. 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)( )n−1
4
2
if n is odd (D) ( )n n − 2
4 if n is even
32. Consider the expansion, (a1 + a2 + a3 +....... + ap)n where n ∈ N and n ≤ p. The correct statement(s) is/are:(A) number of different terms in the expansion is, n + p − 1C n(B) co-efficient of any term in which none of the variables a1, a2 ..., ap occur more than once is ' n '(C) co-efficient of any term in which none of the variables a1, a2,..., ap occur more than once is n ! if
n = p
(D) Number of terms in which none of the variables a1, a2,......, ap occur more than once ispn
.
EXEREXEREXEREXEREXERCISE–5CISE–5CISE–5CISE–5CISE–51. In a telegraph communication how many words can be communicated by using atmost 5 symbols.
(only dot and dash are used as symbols)
2. If all the letters of the word 'AGAIN' are arranged in all possible ways & put in dictionary order, what isthe 50th word.
3. 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.
4. A family consists of a grandfather, m sons and daughters and 2n grand children. They are to be seatedin a row for dinner. The grand children wish to occupy the n seats at each end and the grandfatherrefuses to have a grand children on either side of him. In how many ways can the family be made to sit?
5. The sides AB, BC & CA of a triangle ABC have 3, 4 & 5 interior points respectively on them. Find thenumber of triangles that can be constructed using these interior points as vertices.
6. 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.
7. In how many other ways can the letters of the word MULTIPLE be arranged ; (i) without changing theorder of the vowels (ii) keeping the position of each vowel fixed (iii) without changing the relative order/position of vowels & consonants.
8. There are p intermediate stations on a railway line from one terminus to another. In how many ways cana train stop at 3 of these intermediate stations if no 2 of these stopping stations are to be consecutive?
9. Find the number of positive integral solutions of x + y + z + w = 20 under the following conditions:(i) Zero values of x, y, z, w are include(ii) Zero values are excluded(iii) No variable may exceed 10; Zero values excluded(iv) Each variable is an odd number(v) x, y, z, w have different values (zero excluded).
10. Find the number of words each consisting of 3 consonants & 3 vowels that can be formed from theletters of the word “CIRCUMFERENCE”. In how many of these C’s will be together.
11. If ' n ' distinct things are arranged in a circle, show that the number of ways of selecting three of these
things so that no two of them are next to each other is, 61
n (n − 4) (n − 5).
12. In maths paper there is a question on "Match the column" in which column A contains 6 entries & eachentry of column A corresponds to exactly one of the 6 entries given in column B written randomly.2 marks are awarded for each correct matching & 1 mark is deducted from each incorrect matching.
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20 of
27
A student having no subjective knowledge decides to match all the 6 entries randomly. Find the numberof ways in which he can answer, to get atleast 25 % marks in this question.
13. Show that the number of combinations of n letters together out of 3n letters of which n are a and n areb and the rest unlike is, (n + 2). 2n − 1.
14. Find the number of positive integral solutions of, (i) x2 − y2 = 352706 (ii) xyz = 21600
15. There are ' n ' straight line in a plane, no two of which are parallel and no three pass through the samepoint. Their points of intersection are joined. Show that the number of fresh lines thus introduced is,
81 n (n − 1) (n − 2) (n − 3).
16. A forecast is to be made of the results of five cricket matches, each of which can be a win or a draw ora loss for Indian team. Find(i) number of forecasts with exactly 1 error(ii) number of forecasts with exactly 3 errors(iii) number of forecasts with all five errors
17. Prove by permutation or otherwise ( )( )n
2
!n!n
is an integer (n ∈ I+). [IIT – 2004]
18. If total number of runs scored in n matches is
+
41n
(2n+1 – n – 2) where n > 1, and the rund scored
in the kth match are given by k. 2n+1–k, where 1 ≤ k ≤ n. Find n [IIT – 2005]
ANSWER KEY
EXEREXEREXEREXEREXERCISE–1CISE–1CISE–1CISE–1CISE–1Q.1 m+n+kC3 − (mC3 + nC3 + kC3) Q.2 744 Q.3 13 , 156
Q.4 4316527 Q.5 43200 Q.6 145 Q.7 420
Q.8 960 Q.9 24C2 . 15C3 Q.10 1106 Q.11 12! ; !2)!3(
!4.!114
Q.13 4)!13(!52 ; 3)!17(!3
!52 Q.14 5400 Q.15 45C6 Q.17 1638
Q.18 22100 , 52 Q.19 2111 Q.20 576
Q.21 (a) 5 · (6!) , (b) 3! · 4!, (c) 12 Q.23 (a) 72 ; 78120 ; (b) 23 ; (c) 32
Q.24 143 Q.25 440 Q.26 n + 5C5 Q.27 3888
Q.28!9!5!)14( Q.29 (ii) 792 ; (iii) 4025
143 ; (v) r = 3
EXEREXEREXEREXEREXERCISE–2CISE–2CISE–2CISE–2CISE–2Q.1 2500 Q.2 mCk−1 where m = (1/2) (2n − k² + k − 2)
Q.3 710 ; 496
10 ! Q.4 24
Q.5 532770 Q.6 (i) 15, (ii) 126, (iii) 60, (iv) 105 Q.7 26250
Q.9 120, 216, 210 Q.10 2252 Q.11 (i) 150 ; (ii) 6 ; (iii) 25 ; (iv) 2 ; (v) 6
Q.12 (i) 3359 ; (ii) 59 ; (iii) 359 Q.13 61, 75 Q.14 10, 3
Q.15 121 Q.17 510 Q.18 2 ! 3 ! 8 !
Q.1924
3 2!)2 5
!( !) ( ; 8C1 .
233 2!)2 5
!( !) ( Q.20 ( ) !
( !) ( !) .25
3 4 43 4 Q.21 4 . (4!)² . 8C4 . 6C2
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07
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21 of
27Q.22 240 , 15552 Q.23 3119976 Q.24 27
Q.25 (a) 1680; (b) 1140 Q.26 6C3[3n – 3C1(2
n – 2) – 3C2]
Q.27 974 Q.29 1506
EXEREXEREXEREXEREXERCISE–3CISE–3CISE–3CISE–3CISE–3Q.1 72 Q.2 (i) A ; (ii) B Q.3 C Q.4 B
Q.5 A Q.6 B Q.8 C Q.9 C
EXEREXEREXEREXEREXERCISE–4CISE–4CISE–4CISE–4CISE–41. A 2. C 3. D 4. D 5. D 6. C 7. D
8. D 9. C 10. C 11. B 12. D 13. D 14. D
15. A 16. A 17. C 18. B 19. B 20. B 21. C
22. C 23. BCD 24. CD 25. AB 26. AD
27. CD 28. AB 29. ABC 30. AD 31. CD 32. ACD
EXEREXEREXEREXEREXERCISE–5CISE–5CISE–5CISE–5CISE–51. 62 2. NAAIG 3. 154
4. (2n)! m! (m − 1) 5. 205 6. 744
7. (i) 3359 (ii) 59 (iii) 359
8. p – 2C3
9. (i) 23C3 (ii) 19C3 (iii) 19C3 – 4.9C3 (iv) 11C8 (v) 552
10. 22100, 52 12. 56 ways
14. (i) Zero (ii) 1260 16. (i) 10 (ii) 80 (iii) 32
18. 7
