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rom
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 ABC, ACB, BCA, BAC, CAB and CBA are different arrange- (iii) Problems based on seated in a row
ments of three letters A, B and C, because in each arrangement, (iv) Problems based on digits
order in which the letters arranged, is considered. But if the order (v) Problems based on arrangement letters of a word
in which the things are arranged is not considered; then ABC, (vi) Problems based on rank of a word (in a dictionary)
ACB, BCA, BAC, CAB and CBA are not different but the same.
Problems of Combinations
Similarly AB and BA are not different but the same.
Each of the different selections or groups which can be made (i) Problems based on selections or choose
by some or all of a number of given things without reference to the (ii) Problems based on groups or committee
order of things in any selection or group is called a combination. (iii) Problems based on geometry
As in selection order in which things are selected is not If in any problem, it is neither mentioned that the problem is
considered; hence, selections of two letters AB and BA out of of permutation or combination nor does the problem fall in the
three letters A, B and C are the same. Similarly selections of BC categories mentioned above for the problems of permutations or
and CB are the same. problems of combinations, then do you think whether arrangement
Also selections of CA and AC are the same. (i.e. order) is meaningful or not? If arrangement (i.e., order)
Hence selection of two letters out of the three letters A, B and is considerable in the given problem, then the problem is of
C can be made as AB, BC and CA only. permutation otherwise it is of combination. This will be more
As in arrangements, order in which things are arranged is clear through the following illustrations:
considered. Hence all arrangements of two letters out of the three

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Suppose you have to select three batsmen out of four batsmen
letters A, B and C are AB, BA, BC, CB, CA and AB.
B1, B2, B3 and B4, you can select three batsmen B1 B2 B3, B2 B3
B4, B3 B4 B1 or B4 B1 B2.
Here order of selections of three batsmen in any group of

m three batsmen is not considerable because it does not make any

difference in the match.
Hence in the selection process; B2 B3 B4, B2 B4 B3, B3 B2 B4,
B3 B4 B2, B4 B2 B3 and B4 B3 B2 all are the same.
But for batting, the order of batting is important.
Therefore for batting; B2 B3 B4, B2 B4 B3, B3 B2 B4, B3 B4 B2, B4
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B2 B3 and B4 B3 B2, are different because B2 B3 B4 means batsman
Number of permutations (or arrangements) of two letters out B2 batting first then batsman B3 and then batsman B4 whereas B2
of three letters A, B and C = 6. B4 B3 means batsman B2 batting first then batsman B4 and then
Number of combinations (or groups) of two letters out of three batsman B3.
letters A, B and C = 3.
Permutations of three different letters A, B and C taken two COUNTING FORMULA FOR LINEAR
at a time is also understood as selections of any two different PERMUTATIONS
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letters AB, BC or CA out of A, B and C, then the selected two

Without Repetition
letters arranged in two ways as
AB, BA ; BC, CB or CA, AC 1. Number of permutations of n different things, taking r at a
n
Hence using multiplication principle, number of permutations time is denoted by Pr or P(n, r), which is given by
of three different letters A, B and C taken two at a time n n!
= (Number of ways to select any two different letters out of the Pr = (0 ≤ r ≤ n)
(n − r )!
three given letters) × (Number of arrangements of two selected
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letters) = n(n – 1) (n – 2) ....... (n – r + 1),

=3×2=6 where n is a natural number and r is a whole number.
2. Number of arrangements of n different objects taken all at
Thus permutations means selection of some or all of the given n
things at a time and then arrangements of selected things. In most a time is Pn = n !
of the problems, it is mentioned that the problem is of permutation Note:
or combination but in some problems it is not mentioned. In the
case where it is not mentioned that problem given is of permuta-
n
P1 = n, n
Pr = n. n −1Pr −1 , n
Pr = (n − r + 1). n Pr −1 ,
tion or combination, you can easily identify the given problem is n n
Pn = Pn −1
of permutation or combination using the following classifications
of problems:
Illustration 6: Find the number of ways in which four persons
Problems of Permutations can sit on six chairs.
(i) Problems based on arrangements Solution: 6P4 = 6.5.4.3 = 360
(ii) Problems based on standing in a line
With Repetition CIRCULAR PERMUTATIONS
1. Number of permutations of n things taken all at a time, if 1. Arrangement Around a Circular Table
out of n things p are alike of one kind, q are alike of second
In circular arrangements, there is no concept of starting point
kind, r are alike of a third kind and the rest n – (p + q + r)
(i.e. starting point is not defined). Hence number of circular
are all different is
permutations of n different things taken all at a time is (n – 1)!
n!
if clockwise and anti-clockwise order are taken as different.
p! q! r !
2. Number of permutations of n different things taken r at a
time when each thing may be repeated any number of times
is n r.
Illustration 7: Find the number of words that can be formed
out of the letters of the word COMMITTEE taken all at a time.
Solution: There are 9 letters in the given word in which two T’s,
two M’s and two E’s are identical. Hence the required number of
9! 9! 9! In the case of four persons A, B, C and D sitting around a
words = = = = 45360
2! 2! 2! (2!) 3
8 circular table, then the two arrangements ABCD (in clock-
wise direction) and ADCB (the same order but in anti-

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clockwise direction) are different.
NUMBER OF LINEAR PERMUTATIONS UNDER Hence the number of arrangements (or ways) in which four
CERTAIN CONDITIONS different persons can sit around a circular table = (4 – 1)!
1. Number of permutations of n different things taken all = 3! = 6.

m
together when r particular things are to be placed at some r
given places = n – rPn – r = (n – r)!
2. Number of permutations of n different things taken r at a
time when m particular things are to be placed at m given
places = n – mPr – m
3. Number of permutations of n different things, taken r at a
2. Arrangement of Beads or Flowers (All

Different) Around a Circular Necklace or

Garland
The number of circular permutations of n different things
taken all at a time is
(n − 1)!
2
, if clockwise and anti-clockwise
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time, when a particular thing is to be always included in each order are taken as the same.
arrangement, is r . n – 1Pr – 1
If we consider the circular arrangement, if necklace made
4. Number of permutation of n different things, taken r at a time, of four precious stones A, B, C and D; the two arrangements
when m particular thing is never taken in each arrangement ABCD (in clockwise direction) and ADCB (the same but in
is n – mPr. anti-clockwise direction) are the same because when we take
5. Number of permutations of n different things, taken all at a one arrangement ABCD (in clockwise direction) and then
time, when m specified things always come together is turn the necklace around (front to back), then we get the
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m! × (n – m + 1)! arrangement ADCB (the same but in anti-clockwise direc-

6. Number of permutations of n different things, taken all at a tion). Hence the two arrangements will be considered as one
time, when m specific things never come together is arrangement because the order of the stones is not
n! – m! × (n – m + 1)! changing with the change in the side of observation. So in
Illustration 8: How many different words can be formed this case, there is no difference between the clockwise and
with the letters of the word ‘JAIPUR’ which start with ‘A’ anti-clockwise arrangements.
and end with ‘I ’? Therefore number of arrangements of four different stones
Solution: After putting A and I at their respective places (only (n − 1)!
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in the necklace = .
in one way) we shall arrange the remaining 4 different letters at 2
4 places in 4! ways. Hence the required number = 1 × 4! = 24.
3. Number of Circular Permutations of n
Illustration 9: How many different 3 letter words can be
Different Things Taken r at a Time

formed with the letters of word ‘JAIPUR’ when A and I are
always to be excluded?
Case I: If clockwise and anti-clockwise orders are taken as
Solution: After leaving A and I, we are remained with 4 different, then the required number of circular permutations
n
different letters which are to be used for forming 3 letters words. Pr
= .
Hence the required number = 4P3 = 4 × 3 × 2 = 24. r

Case II: If clockwise and anti-clockwise orders are taken
as same, then the required number of circular permutations
n
Pr
= .
2r
 4. Restricted Circular Permutations Illustration 15: How many combinations of 4 letters can be
When there is a restriction in a circular permutation then first made of the letters of the word ‘JAIPUR’ ?
of all we shall perform the restricted part of the operation Solution: Here 4 things are to be selected out of 6 different things.
and then perform the remaining part treating it similar to a 6.5.4.3
linear permutation. So the number of combinations = 6C4 = = 15
4.3.2.1
Illustration 10: In how many ways can 5 boys and 5 girls be
seated at a round table so that no two girls may be together ? 2. Selection of Objects With Repetition
Solution: Leaving one seat vacant between two boys, 5 boys may
The total number of selections of r things from n different
be seated in 4! ways. Then at remaining 5 seats, 5 girls can sit in things when each thing may be repeated any number of times
5! ways. Hence the required number = 4! × 5! is n + r – 1Cr
Illustration 11: In how many ways can 4 beads out of 6 different 3. Restricted Selection
beads be strung into a ring ? (i) Number of combinations of n different things taken r at
Solution: In this case a clockwise and corresponding anticlock- a time when k particular things always occur is n – kCr – k.
wise order will give the same circular permutation. So the required (ii) Number of combinations of n different things taken r at
6
P4 6.5.4.3 a time when k particular things never occur is n – kCr.
number == = 45 .
4.2 4.2 4. Selection From Distinct Objects
Illustration 12: Find the number of ways in which 10 persons Number of ways of selecting at least one thing from n

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can sit round a circular table so that none of them has the same different things is
neighbours in any two arrangements. nC + nC + nC + .....+ nC = 2n – 1.

1 2 3 n
Solution: 10 persons can sit round a circular table in 9! ways. This can also be stated as the total number of combination
But here clockwise and anti-clockwise orders will give the same of n different things is 2n – 1.
neighbours. Hence the required number of ways = 9! .

COUNTING FORMULA FOR COMBINATION

1. Selection of Objects Without Repetition

1
2

The number of combinations or selections of n different

m Illustration 16: Ramesh has 6 friends. In how many ways can
he invite one or more of them at a dinner ?
Solution: He can invite one, two, three, four, five or six friends
at the dinner. So total number of ways of his invitation
= 6C1 + 6C2 + 6C4 + 6C5 + 6C6 = 26 – 1 = 63
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things taken r at a time is denoted by nCr or C (n, r) or 5. Selection From Identical Objects
 n (i) The number of combination of n identical things taking
C  
r r (r ≤ n) at a time is 1.
(ii) The number of ways of selecting any number r (0 ≤ r ≤ n)
nC
n!
where r = ; (0 ≤ r ≤ n) of things out of n identical things is n + 1.
r ! (n − r )!
(iii) The number of ways to select one or more things out
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n(n − 1)(n − 2)...(n − r + 1) of (p + q + r) things; where p are alike of first kind, q

= ; are alike of second kind and r are alike of third kind
r (r − 1)(r − 2)....2.1
where n is a natural number and r is a whole number. = (p + 1) (q + 1) (r + 1) – 1.
Some Important Results Illustration 17: There are n different books and p copies of
n
Pr each in a library. Find the number of ways in which one or
(i) nCn = 1 , nC0 = 1 (ii) nCr = more than one books can be selected.
r!
(iii) nCr = nCn – r (iv) nCx = nCy ⇒ x + y = n Solution: Required number of ways
= (p + 1)(p +1)......n terms – 1 = (p + 1)n – 1
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n n–1
(v) nCr + nCr – 1= n + 1Cr (vi) nCr = . Cr – 1 Illustration 18: A bag contains 3 one ` coins, 4 five ` coins
r
(vii) nC1 = nCn – 1 = n and 5 ten ` coins. How many selection of coins can be formed
by taking atleast one coin from the bag?
Illustration 13: If 20C = 20Cr – 10, then find the value of 18Cr
r Solution: There are 3 things of first kind, 4 things of second
Solution: 20C
r = 20Cr – 10 ⇒ r + (r – 10) = 20 ⇒ r = 15 kind and 5 things of third kind, so the total number of selections
18.17.16 = (3 + 1) (4 + 1) (5 + 1) – 1 = 119
∴  18Cr = 18C15 = 18C3 = = 816
1.2.3
Illustration 14: How many different 4-letter words can be
DIVISION AND DISTRIBUTION OF OBJECTS
formed with the letters of the word ‘JAIPUR’ when A and I
are always to be included ? 1. The number of ways in which (m + n) different things can
Solution: Since A and I are always to be included, so first we be divided into two groups which contain m and n things
select 2 letters from the remaining 4, which can be done in respectively is
4C = 6 ways. Now these 4 letters can be arranged in 4! = 24 ways, m + nC nC =
(m + n)!
2 m n ,m≠n
so the required number = 6 × 24 = 144. m !n !

Particular case: (ii) Number of triangles obtained by joining these n points
When m = n, then total number of ways is = nC3
(2m)! (iii) Number of quadrilaterals obtained by joining these n
, when order of groups is considered and
(m !) 2 points = nC4
(2m)! Illustration 21: There are 10 points in a plane and 4 of them
, when order of groups is not considered. are collinear. Find the number of straight lines joining any
2!(m !) 2 two of them.
2. The number of ways in which (m + n + p) different things Solution: Total number of lines = 10C2 – 4C2 + 1 = 40.
can be divided into three groups which contain m, n and p Illustration 22: If 5 parallel straight lines are intersected by 4
things respectively is parallel straight lines, then find the number of parallelograms
m + n + pC . n + pC . pC =
(m + n + p )! thus formed.
m p p ,m≠n≠p Solution:
m !n ! p !

Particular case:
When m = n = p, then total number of ways is
(3m)!
, when order of groups is considered and
(m !)3
(3m)!
, when order of groups is not considered.

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3!(m !)3
3. (i) Total number of ways to divide n identical things among Number of parallelograms = 5C2 × 4C2 = 60.
r person isn + r – 1C
r–1
(ii) Also total number of ways to divide n identical things

Illustration 19: In how many ways 20 identical mangoes may

be divided among 4 persons if each person is to be given at
least one mango?
m
among r persons so that each gets atleast one is n – 1Cr – 1.

Solution:  If each person is to be given at least one mango, then

number of ways will be 20 – 1C4 – 1 = 19C3 = 969.
FINDING THE RANK OF A WORD
We can find the rank of a word out of all the words with or
without meaning formed by arranging all the letters of a given
word in all possible ways when these words are listed as in a
dictionary. You can easily understand the method to find the above
mentioned rank by the following illustrations.
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Illustration 20: In how many ways can a pack of 52 cards Illustration 23: If the letters of the word RACHIT are
be divided in 4 sets, three of them having 17 cards each and arranged in all possible ways and these words (with or without
fourth just one card? meaning) are written as in a dictionary, then find the rank of
Solution: Since the cards are to be divided into 4 sets, 3 of them this word RACHIT.
having 17 cards each and 4th just one card, so number of ways Solution: The order of the alphabet of RACHIT is A, C, H, I, R, T.
52! 51! 52! The number of words beginning with A (i.e. the number of
= . = . words in which A comes at first place) is 5P5 = 5!.
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1!51! (17!)3 3! (17!)3 3!

Similarly, number of words beginning with C is 5!, beginning
with H is 5! and beginning with I is also 5!.
IMPORTANT RESULTS ABOUT POINTS So before R, four letters A, C, H, I can occur in 4 × (5!) = 480
1. If there are n points in a plane of which m ( < n) are collinear, ways.
then Now the word RACHIT happens to be the first word beginning
(i) Total number of different straight lines obtained by with R. Therefore the rank of this word RACHIT = 480 + 1 = 481.
joining these n points is nC2 – mC2 + 1. Illustration 24: The letters of the word MODESTY are
(ii) Total number of different triangles formed by joining written in all possible orders and these words (with or without
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these n points is nC3– mC3 meaning) are listed as in a dictionary then find the rank of
2. Number of diagonals of a polygon of n sides is nC2 – n i.e., the word MODESTY.
n (n − 3) Solution:
. The order of the alphabet of MODESTY is D, E, M, O, S, T, Y.
2
Number of words beginning with D is 6P6 = 6!
3. If m parallel lines in a plane are intersected by a family of Number of words beginning with E is 6P6 = 6!
other n parallel lines, then total number of parallelograms
Number of words beginning with MD is 5P5 = 5!
mn (m − 1) (n − 1) Number of words beginning with ME is 5P5 = 5!
so formed is mC2 × nC2 i.e., .
4 Now the first word start with MO is MODESTY.
Hence rank of the word MODESTY
4. Given n points on the circumference of a circle, then = 6! + 6! + 5! + 5! + 5! + 1
(i) Number of straight lines obtained by joining these n = 720 + 720 + 120 + 120 + 1
points = nC2 = 1681.
jk
ch
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m
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19. If a secretary and a joint secretary are to be selected from a 30. The number of ways in which 7 different books can be given
committee of 11 members, then in how many ways can they to 5 students if each can receive none, one or more books is
be selected? (a) 57 (b) 75
(a) 110 (b) 55 (c) 11C5 (d) 12!
(c) 22 (d) 11 31. In how many ways can 13 different alphabets (a, b, c, ... m)
20. On a railway route there are 20 stations. What is the number be arranged so that the alphabets f and g never come together?
of different tickets required in order that it may be possible
(a) 13 ! – 12 ! (b) 13 ! – 12! / 2!
to travel from every station to every other station?
(c) 13 ! – 2 × 12 ! (d) None of these
(a) 40 (b) 380
(c) 400 (d) 420 32. Number of ways in which the letters of word GARDEN
can be arranged with vowels in alphabetical order, is
21. If P(32, 6) = kC (32, 6), then what is the value of k?
(a) 6 (b) 32 (a) 360 (b) 240
(c) 120 (d) 720 (c) 120 (d) 480
22. How many straight lines can be formed from 8 non-collinear 33. If 5 parallel straight lines are intersected by 4 parallel
points on the X-Y plane? straight lines, then the number of parallelograms thus formed
(a) 28 (b) 56 is
(c) 18 (d) 19860 (a) 20 (b) 60

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23. A man has 3 shirts, 4 trousers and 6 ties. What are the number (c) 101 (d) 126
of ways in which he can dress himself with a combination 34. The number of ways in which a couple can sit around a
of all the three? table with 6 guests if the couple take consecutive seat is
(a) 13 (b) 72 (a) 1440 (b) 720
(c) 13!/3!×4!×6!

(a) 10
(c) 7
(d) 3!×4!×6!

(b) 11
(d) 9
m
24. If (28C2r : 24C2r–4) = 225 : 11. Find the value of r.

25. How many numbers can be formed with the digits 1, 6, 7, 8,

6, 1 so that the odd digits always occupy the odd places.
35.
(c) 5040 (d) None of these
How many different words beginning with O and ending
with E can be formed with the letters of the word
ORDINATE, so that the words are beginning with O and
ending with E?
ro
(a) 8! (b) 6!
(a) 15 (b) 12 (c) 7! (d) 7!/2!
(c) 18 (d) 20 36. How many 6 digit number can be formed from the digits 1,
26. There are 20 people among whom two are sisters. Find the 2, 3, 4, 5, 6 which are divisible by 4 and digits are not
number of ways in which we can arrange them around a repeated?
circle so that there is exactly one person between the two
(a) 192 (b) 122
sisters.
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(c) 140 (d) 242

(a) 18! (b) 2!19!
(c) 19! (d) None of these 37. In how many ways can the eight directors, the vice-
chairman and the chairman of a firm be seated at a round-
27. In a company, each employee gives a gift to every other
employee. If the number of gifts is 61, then the number of table, if the chairman has to sit between the vice-chairman
employees in the company is : and the director?
(a) 11 (b) 13 (a) 9! × 2 (b) 2 × 8!
(c) 12 (d) 8 (c) 2 × 7! (d) None of these
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28. There are three rooms in a hotel: one single, one double 38. In how many different ways can the letters of the word
and one for four persons. How many ways are there to house ‘CREAM’ be arranged ? [SBI Clerk-June-2012]
seven persons in these rooms? (a) 720 (b) 240
(a) 7!/1!2!4! (b) 7! (c) 360 (d) 504
(c) 7!/3 (d) 7!/3! (e) None of these
29. The digits, from 0 to 9 are written on 10 slips of paper 39. Which of the following words can be written in 120 different
(one digit on each slip) and placed in a box. If three of the ways? [IBPS Clerk-2012]
slips are drawn and arranged, then the number of possible
(a) STABLE (b) STILL
different arrangements is
(c) WATER (d) NOD
(a) 1000 (b) 720
(e) DARE
(c) 810 (d) None of these
 Level- II
1. 5 men and 6 women have to be seated in a straight row so 11. If a team of four persons is to be selected from 8 males and
that no two women are together. Find the number of ways 8 females, then in how many ways can the selections be
this can be done. made to include at least one male.
(a) 48400 (b) 39600 (a) 1550 (b) 1675
(c) 9900 (d) 86400 (c) 1725 (d) 1750
2. The total number of ways in which 8 men and 6 women can 12. Letters of the word DIRECTOR are arranged in such a way
be arranged in a line so that no 2 women are together is that all the vowels come together. Find out the total number
(a) 48 (b) 8P8.9P6 of ways for making such arrangement.
(c) 8! (84) (d) 8C8.9C8 (a) 4320 (b) 2720
3. The number of different ways in which 8 persons can stand (c) 2160 (d) 1120
in a row so that between two particular person A and B there 13. 4 boys and 2 girls are to be seated in a row in such a way
are always two person, is that the two girls are always together. In how many different
(a) 60 (5!) (b) 15(4!) × (5!) ways can they be seated?
(c) 4! × 5! (d) None of these (a) 1200 (b) 7200

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4. From 6 boys and 7 girls a committee of 5 is to be formed so (c) 148 (d) 240
as to include atleast one girl. The number of ways this can
14. In how many ways can 7 Englishmen and 7 Americans sit
be done is
down at a round table, no 2 Americans being in consecutive
(a) 13C4 (b) 6C4 . 7C1

5.
6
(c) 7 . C4 (d) 13C5 – 6C1
m
How many different nine digit numbers can be formed from
the number 223355888 by rearranging its digits so that the
odd digits occupy even positions?
(a) 16 (b) 36
positions?
(a) 3628800
(c) 3628000
(b) 2628800
(d) 3328800
15. How many numbers greater than one million can be formed
with 2, 3, 0, 3, 4, 2, 3? (repetitions not allowed)
(a) 720 (b) 360
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(c) 60 (d) 180
(c) 120 (d) 240
6. If two dices are tossed simultaneously, the number of
elements in the resulting sample space is 16. 5 Indian and 5 American couples meet at a party & shake
hands . If no wife shakes hands with her husband and no
(a) 6 (b) 8
Indian wife shakes hands with a male, then the number of
(c) 36 (d) 24
hand shakes that takes place in the party is
7. In how many ways can 7 persons stand in the form of a
ch

(a) 95 (b) 110

ring?
(a) P (7, 2) (b) 7 ! (c) 135 (d) 150
17. The total number of ways in which letters of the word
7!
(c) 6 ! (d) ACCOST can be arranged so that the two C's never come
2
together will be
8. In a football championship 153 matches were played. Every
team played one match with each other team. How many (a) 120 (b) 360
teams participated in the championship? (c) 240 (d) 6 ! – 2 !
jk

(a) 21 (b) 18 18. In how many ways can a term of 11 cricketers be chosen
(c) 17 (d) 15 from 6 bowlers. 4 wicket keepers and 11 batsmen to give a
9. If P(77, 31) = x and C (77, 31) = y, then which one of the majority of bastemen if at least 4 bowlers are to be included
following is correct? and there is one wicket keeper?
(a) x = y (b) 2x = y (a) 27730 (b) 27720
(c) 77x = 31 y (d) x > y (c) 17720 (d) 26720
10. In how many ways can 12 papers be arranged if the best 19. Three dice are rolled. The number of possible outcomes in
and the worst paper never come together? which at least one die shows 5 is
(a) 12!/2! (b) 12! – 11! (a) 215 (b) 36
(c) (12! – 11!)/2 (d) 12! – 2.11! (c) 125 (d) 91
20. The sides AB, BC, CA of a traingle ABC have 3, 4 and 5 30. There are 10 points on a line and 11 points on another line,
interior points respectively on them. The total number of which are parallel to each other. How many triangles can
triangles that can be constructed by using these points as be drawn taking the vertices on any of the line?
vertices is (a) 1,050 (b) 2,550
(a) 220 (b) 204 (c) 150 (d) 1,045
(c) 205 (d) 195 31. How many motor vehicle registration number plates can be
21. If all permutations of the letters of the word AGAIN are formed with the digits 1, 2, 3, 4, 5 (No digits being repeated)
arranged as in dictionary, then fiftieth word is if it is given that registration number can have 1 to 5 digits?
(a) NAAGI (b) NAGAI (a) 100 (b) 120
(c) NAAIG (d) NAIAG (c) 325 (d) 205
22. All the words that can be formed using alphabets A, H, L, U 32. How many different 9-digit numbers can be formed from
and R are written as in a dictionary (no alphabet is repeated). the number 223355888 by rearranging its digits so that the
Rank of the word RAHUL is odd digits occupy even positions?
(a) 71 (b) 72 (a) 120 (b) 9!(2!)3.3!
(c) 73 (d) 74 (c) (4!)(2!)3.(3!) (d) None of these
23. How many new words are possible from the letters of the 33. There are 5 different Jeffrey Archer books, 3 different Sidney
word PERMUTATION? Sheldon books and 6 different John Grisham books. The
number of ways in which at least one book can be given

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(a) 11!/2! (b) (11!/2!) – 1
(c) 11! – 1 (d) None of these away is
24. There are five boys and three girls who are sitting together (a) 210 – 1 (b) 211 –1
to discuss a management problem at a round table. In how (c) 212 – 1 (d) 214 – 1
many ways can they sit around the table so that no two girls 34. How many natural numbers not more than 4300 can be
are together?
(a) 1220
(c) 1420
(b) 1400
(d) 1440
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25. How many 6-digit numbers have all three digits either all
odd or all even? 35.
formed with the digits 0, 1, 2, 3, 4 (if repetitions are
allowed)?
(a) 574
(c) 575
(b) 570
(d) 569
The sides of a triangle have 4, 5 and 6 interior points marked
on them respectively. The total number of triangles that can
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(a) 31,250 (b) 28,125
(c) 15,625 (d) None of these be formed using any of these points
26. Out of 10 consonants and four vowels, the number of words (a) 371 (b) 415
that can be formed using six consonants and three vowels (c) 286 (d) 421
is 36. Total number of ways in which six '+' and four '–' sings can
(a) 10P6 × 6P3 (b) 10C6 × 6C3 be arranged in a line such that no two '–' sings occur together,
(c) 10 4
C6 × C3 × 9! (d) 10P6 × 4P3 is
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27. The number of 5 digit numbers that can be made using the (a) 35 (b) 18
digits 1 and 2 and in which at least one digit is different, is (c) 15 (d) 42
(a) 30 (b) 31 37. In how many ways can 5 prizes be distributed among 4
(c) 32 (d) None of these boys when every boy can take one or more prizes?
28. A class photograph has to be taken. The front row consists (a) 1024 (b) 625
of 6 girls who are sitting. 20 boys are standing behind. The (c) 120 (d) 600
two corner positions are reserved for the 2 tallest boys. In 38. Three dice are rolled. The number of possible outcomes in
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how many ways can the students be arranged? which at least one die shows 5 is
(a) 18! × 1440 (b) 6! × 1440 (a) 215 (b) 36
(c) 18! × 2! × 1440 (d) None of these (c) 125 (d) 91
29. A,B,C and D are four towns any three of which are non- 39. Find the number of numbers between 300 and 3000 that
colinear. Then the number of ways to construct three roads can be formed with the digits 0, 1, 2, 3, 4 and 5, no digit
each joining a pair of towns so that the roads do not form a being repeated. [SBI PO-2011]
triangle is (a) 120 (b) 160
(a) 7 (b) 8 (c) 240 (d) 60
(c) 9 (d) More than 9 (e) None of these
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22. (a) For a straight line we just need to select 2 points out of 37. (b) Let the vice-chairman and the chairman from 1 unit
the 8 points available. 8C2 would be the number of along with the eight directors, we now have to arrange
ways of doing this. 9 different units in a circle.
23. (b) 3C1 × 4C1 × 6C1 = 72 This can be done in 8! ways.
24. (c) At r = 7, the value becomes At the same time, the vice-chairman & the chairman
(28!/14! × 14!) /(24!/10! × 14!) ® 225 : 11 can be arranged in two different ways. Therefore, the
total number of ways = 2 × 8!.
25. (c) The digits are 1, 6, 7, 8, 7, 6, 1. In this seven-digit no.
there are four odd places and three even places 38. (e) CR EAM
1 2 3 4 5
OEOEOEO. The four odd digits 1, 7, 7, 1 can be
arranged in four odd places in [4!/2!×2] = 6 ways [as 1 Required number of ways = 5!
and 7 are both occurring twice]. = 5 × 4 × 3 × 2 × 1 = 120
The even digits 6, 8, 6 can be arranged in three even 39. (c) (a) The word STABLE has six distinct letters.
places in 3!/2! = 3 ways. \ Number of arrangements = 6 !
Total no. of ways = 6 × 3 = 18 = 6 × 5 × 4 × 3 × 2 × 1 = 720
26. (d) First arrange the two sisters around a circle in such a (b) The word STILL has five letters in which letter ‘L’
way that there will be one seat vacant between them. comes twice.
[This can be done in 2! ways since the arrangement of \ Number of arrangements

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the sisters is not circular.] 5!
Then, the other 18 people can be arranged on 18 seats = = 60
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in 18! ways. (c) The word WATER has five distinct letters.
27. (c) Let the total number of employees in the company be n. \ Number of arrangements = 5 ! = 5 × 4 × 3 × 2× 1 = 120
n
Total number of gifts = C2 =
n(n - 1)
2

Þ n 2 - n - 132 = 0 or (n + 11)(n - 12) = 0

or n = 12 [– 11 is rejected]
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= 61 \
(d) The word ‘NOD’ has 3 distinct letters.
Number of arrangements = 3 ! = 6
(e) Number of arrangements = 4 ! = 24

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28. (a) Choose 1 person for the single room & from the 1. (d) Total seats = 5 + 6 = 11.
remaining choose 2 people for the double room & from Arrangement will be : W M W M W M W M W M W
the remaining choose 4 people for the 4 persons room Þ Total possible arrangements will be :
® 7C1 × 6C2 × 4C4. 6P × 5P = 86400.
6 5
29. (b) 10P = 720
3 2. (b) 8 men can sit in a row in 8P8 ways. Then for the 6
30. (a) Ist book can be given to any of the five students. women, there are 9 seats to sit
Similarly other six books also have 5 choices. Hence \ the women can sit in 9P6 ways
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the total number of ways is 57. \ total number of ways = 8P8 . 9P6
31. (c) Total possible arrangements = 13P13 = 13! 3. (a) The number of 4 persons including A, B = 6C2
Total number in which f and g are together Considering these four as a group, number of
= 2 × 12P12 = 2 × 12! arrangements with the other four = 5!
32. (a) Order of vowels of fixed But in each group the number of arrangements
6! = 2! × 2!
\ required number of ways are \ The required number of ways = 6C2 × 5! × 2! × 2!
2!
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33. (b) Number of parallelograms = 5C2 × 4C2 = 60. 4. (d) From total 13 members 5 can be select as 13C5
34. (a) A couple and 6 guests can be arranged in (7 – 1) ways. For at least one girl in the committee, number of ways
But in two people forming the couple can be arranged are 13C5 – 6C1
among themselves in 2! ways. 5. (c) X - X - X - X - X. The four digits 3, 3, 5,5 can be arranged
\ the required number of ways = 6! × 2! = 1440 4!
at (–) places in = 6 ways.
35. (b) 6! ways, O fixed 1st and E fixed in last. 2!2!
36. (a) For the number to be divisible by 4, the last two digits The five digits 2, 2, 8, 8, 8 can be arranged at (X) places
must be any of 12, 24, 16, 64, 32, 36, 56 and 52. The 5!
last two digit places can be filled in 8 ways. Remaining in ways = 10 ways
2!3!
3 places in 4P3 ways. Hence no. of 5 digit nos. which
Total no. of arrangements = 6 ´ 10 = 60 ways
are divisible by 4 are 24 × 8 = 192.
6. (c) Number of elements in the sample space 14. (a) Putting l Englishman in a fixed position, the remaining
= 6 × 6 = 36 6 can be arranged in 6! 720 ways, For each such
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) arrangement, there are 7 positions for the 7 Americans
7. (c) Number of ways in which 7 persons can stand in the and they can be arranged in 7! ways.
form of a ring = (7 – 1) ! = 6! Total number of arrangements = 7! × 6! = 3628800
8. (b) Let total no. of team participated in a championship be 15. (b) Required number is greater than 1 million (7 digits).
n. From given digits, total numbers which can be formed
Since, every team played one match with each other = 7!
team. Number starting from zero = 6!
n! Þ Required number = 7! – 6!
\ nC2 = 153 Þ = 153
2!( n – 2)! Q Repetition not allowed, so required answer

n(n – 1)(n – 2)! n(n –1) 7! - 6!

= 153 Þ = 153 = = 360
Þ 2!3!
2!(n – 2)! 2
Þ n(n – 1) = 306 16. (c) Total number of hand shakes = 20C2 of those no Indian
female shakes hand with male
Þ n2 – n – 306 = 0

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Þ 5 × 10 = 50 hand shakes
Þ n2 – 18n + 17n – 306 = 0
No American wife shakes hand with her husband
Þ n (n – 18) + 17 (n – 18) = 0
Þ n = 18, – 17 = 5 × 1 = 5 hand shakes
Þ total number of hand shakes occurred

9. (d)
n cannot be negative
\ n ¹ – 17
Þ n = 18
As we know
P(n, r) = r! C (n, r)
\ From the question, we have
m = 20C2 – (50 + 5) = 190 – 55 = 135
17. (c) Total number of ways to permute 6 alphabets 2 of which
are common = 6! / 2! = 360.
(1) Treat the two C’s as one
Þ Number of possible ways = 5P5 = 120
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x = r ! (y) (b) Number of ways = Total arrangements – Number
of arrangements in which they always come together
Here r = 31
= 360 – 120 = 240.
\ x = (31)!. y.
18. (b) 1 wicket keeper from 4 can be selected in
10. (d) All arrangements – Arrangements with best and worst
paper together = 12! – 2! × 11!. 4!
4C = = 4 ways
1 m + 3f = 8C1 × 8C3 = 8 × 56 = 448
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11. (d) 1
3!.1!
2 m + 2f = 8C2 × 8C2 = 28 × 28 = 784
If 4 bowlers are chosen then remaining 6 batsmen -
3 m + 1f = 8C3 × 8C1 = 56 × 8 = 448 can be chosen in 11C6.
4 m + 8f = 8C4 × 8C0 = 70 × 1 = 70
Total = 1750 6C . 11C =
6! 11! 5 ´ 6 11´10 ´ 9 ´ 8 ´ 7
4 6 × = ×
12. (c) Taking all vowels (IEO) as a single letter (since they 4!.2! 3!.1! 2 5´ 4´ 3´ 2
come together) there are six letters among which there = 15 × 14 × 33 = 6930
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are two R.
If we choose 5 bowlers then we have to choose 5
6! batsmen
Hence no. of arrangements = × 3! = 2160
2!
\ there is no majority.
There vowels can be arranged in 3! ways among
themselves, hence multiplied with 3!. \ Total number of ways = 4 × 6930 = 27720.
13. (d) Assume the 2 given students to be together (i.e. one). 19. (d) Required number of possible outcomes
Now these are five students. = Total number of possible outcomes –
Possible ways of arranging them are = 5! = 120 Number of possible outcomes in which 5 does
Now they (two girls) can arrange themselves in 2! not appear on any dice. (hence 5 possibilities in each
ways. throw)
Hence total ways = 120 × 2 = 240 = 63 – 53 = 216 – 125 = 91
20. (c) We have in all 12 points. Since, 3 points are used to 27. (a) Total number of numbers without restriction = 25
form a traingle, therefore the total number of traingles Two numbers have all the digits equal. So, the required
including the triangles formed by collinear points on numbers = 25 – 2 = 30.
AB, BC and CA is 12C3 = 220. But this includes the
28. (a) Two tallest boys can be arranged in 2! ways. Rest 18
following :
can be arranged in 18! ways.
The number of traingles formed by 3 points on
Girls can be arranged in 6! ways.
AB =3C3 = 1
Total number of ways of arrangement = 2! × 18! × 6!
The number of triangles formed by 4 points on
= 18! × 2 × 720 = 18! × 1440
BC = 4C3 = 4.
29. (d) To construct 2 roads, three towns can be selected out
The number of triangles formed by 5 points on
of 4 in 4 × 3 × 2 = 24 ways.
CA = 5C3 = 10.
Now if the third road goes from the third town to the
Hence, required number of traingles
first town, a triangle is formed, and if it goes to the
= 220 – (10 + 4 + 1) = 205.
fourth town, a triangle is not formed. So, there are 24
21. (c) Starting with the letter A, and arranging the other four
ways to form a triangle and 24 ways of avoiding a
letters, there are 4! = 24 words. These are the first 24
triangle.
words. Then starting with G, and arranging A, A, I, and
30. (d) For a triangle, two points on one line and one on the

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4! 24
N in different ways, there are = = 12 words. other has to be chosen.
2!1!1! 2
10
Hence, total 36 words. No. of ways = C2 × 11C1 + 11C2 × 10 C1 =1,045 .
Next, the 37th word starts with I. There are 12 words
31. (c) Single digit numbers = 5

22. (d)
49th word is NAAGI. The 50th word is NAAIG.
No. of words starting with A are 4 ! = 24
No. of words starting with H are 4 ! = 24
No. of words starting with L are 4 ! = 24
These account for 72 words
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starting with I. This accounts up to the 48th word. The
Two digit numbers = 5 × 4 = 20
Three digit numbers = 5 × 4 × 3 = 60
Four digit numbers = 5 × 4 × 3 × 2 = 120
Five digit numbers = 5 × 4 × 3 × 2 × 1 = 120
Total = 5 + 20 + 60 + 120 + 120 = 325
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Next word is RAHLU and the 74th word RAHUL 32. (d) The odd digits have to occupy even positions. This
23. (b) Number of 11 letter words formed from the letter P, E, 4!
R, M, U, T, A, I, O, N = 11!/2!. can be done in = 6ways
2!2!
Number of new words formed = total words – 1
= 11!/2! – 1. The other digits have to occupy the other positions.
24. (d) We have no girls together, let us first arrange the 5
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5!
boys and after that we can arrange the girls in the space This can be done in = 10 ways
3!2!
between the boys.
Number of ways of arranging the boys around a circle Hence total number of rearrangements possible
= [5 – 1]! = 24. = 6 × 10 = 60.
Number of ways of arranging the girls would be by 33. (d) For each book we have two options, give or not give.
placing them in the 5 spaces that are formed between Thus, we have a total of 214 ways in which the 14 books
the boys. This can be done in 5P3 ways = 60 ways. can be decided upon. Out of this, there would be 1
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Total arrangements = 24 × 60 = 1440. way in which no book would be given. Thus, the
25. (b) When all digits are odd number of ways is 214 – 1.
5 × 5 × 5 × 5 × 5 × 5 = 56 34. (c) The condition is that we have to count the number of
When all digits are even natural numbers not more than 4300.
4 × 5 × 5 × 5 × 5 × 5 × 5 = 4 × 55 The total possible numbers with the given digits
56 + 4 × 55 = 28125 = 5 × 5 × 5 × 5 = 625 – 1 = 624.
26. (c) Six consonants and three vowels can be selected from Subtract from this the number of natural number greater
10 consonants and 4 vowels in 10C6 × 4C3 ways. Now, than 4300 which can be formed from the given digits
these 9 letters can be arranged in 9! ways. So, required = 1 × 2 × 5 × 5 – 1 = 49.
number of words = 10C6 × 4C3 × 9!. Hence, the required number of numbers = 624 – 49.
35. (d) You can form triangles by taking 1 point from each \ the number of ways of their distribution
side, or by taking 2 points from any 1 side and the = 4 × 4 × 4 × 4 × 4 = 45 = 1024.
third point from either of the other two sides. 38. (d) Required number of possible outcomes
This can be done in: 4 × 5 × 6 = 4C2 × 11C1 + 5C2 = Total number of possible outcomes – Number of
× 10C1 + 6C2 × 9C1 = 120 + 66 + 100 + 135 = 421 possible outcomes in which 5 does not appear on any
36. (a) First we write six '+' sings at alternate places i.e., by dice
leaving one place vacant between two successive '+' = 63 – 53 = 91.
sings. Now there are 5 places vacant between these 39. (e) 3 5 4
sings and these are two places vacant at the ends. If we
(3 or 4 or 5)
write 4 '–' sings these 7 places then no two '–' will come
3 × 5 × 4 = 60
together. Hence total number of ways 7C4 = 35
2 5 4 3
37. (a) First prize may be given to any one of the 4 boys, hence
first prize can be distributed in 4 ways. (1 or 2)
Similarly every one of second, third, fourth and fifth 2 × 5 × 4 × 3 = 120
prizes can also be given in 4 ways. Total = 120 + 60 = 180

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