1. science
2. mathematics
3. arithmetic

PERMUTATIONS AND COMBINATIONS
Formulae
1. Factorial Notation:
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n - 1)(n - 2) ... 3.2.1.
Examples:
i. We define 0! = 1.
ii. 4! = (4 x 3 x 2 x 1) = 24.
iii. 5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations:
The different arrangements of a given number of things by taking some or all at a time,
are called permutations.
Examples:
i.
All permutations (or arrangements) made with the letters a, b, c by taking two at a
time are (ab, ba, ac, ca, bc, cb).
ii. All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)
3. Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
n
Pr = n(n - 1)(n - 2) ... (n - r + 1) =
n!
(n  r )!
Examples:
i. 6P2 = (6 x 5) = 30.
ii. 7P3 = (7 x 6 x 5) = 210.
iii. Cor. number of all permutations of n things, taken all at a time = n!.
4. An Important Result:
If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3
are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.
Then, number of permutations of these n objects is =
n!
( p1!)( p 2 )!....( p r !)
5. Combinations:
Each of the different groups or selections which can be formed by taking some or all of a
number of objects is called a combination.
Examples:
1. Suppose we want to select two out of three boys A, B, C. Then, possible
selections are AB, BC and CA.
Note: AB and BA represent the same selection.
2. All the combinations formed by a, b, c taking ab, bc, ca.
3. The only combination that can be formed of three letters a, b, c taken all at a time
is abc.
4. Various groups of 2 out of four persons A, B, C, D are:
AB, AC, AD, BC, BD, CD.
5. Note that ab ba are two different permutations but they represent the same
combination.
Number of Combinations:
The number of all combinations of n things, taken r at a time is:
n
Cr =
n(n  1)(n  2)....to r factprs
n!
=
r!
(r!)(n  r!)
Note:
.
i.
n
Cn = 1 and nC0 = 1.
n
Cr = nC(n - r)
Examples:
i) 11C4 =
11  10  9  8
 330
(4  3  2  1)
ii) 16C13 = 16C(16 - 13) = 16C3 =
16  15  14 16  15  14

 560
3!
3  2 1
1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so
that at least 3 men are there on the committee. In how many ways can it be done?
A)564
B)645
C) 735
D) 756
E) None of these
Answer: Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
 7  6 5 6 5  7  6 5 4 6  7  6 5 4 3
=

  +
 

 3  2 1 2 1   4  3  2 1 1   5  4  3  2 1 
=(35x15)+(35x6)+21
=525+210+21
=756
2. When four fair dice are rolled simultaneously, in how many outcomes will at least one of the
diceshow3?
(1)155
(2)620
(3)671
(4) 625
Correct Answer - (3)
Solution:
When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes.
The number of outcomes in which none of the 4 dice show 3 will be 54 = 625 outcomes.
Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671
3. In how many different ways can the letters of the word 'CORPORATION' be arranged so that
the vowels always come together?
A) 810
B) 1440
C) 2880
D) 50400
E)5760
Answer: Option D
Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =
7!
= 2520
2!
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in
5!
= 20
3!
ways.
Required number of ways = (2520 x 20) = 50400.
4. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be
formed?
A)210
B)1050
C) 25200
D) 21400
E) None of these
Answer: Option C
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= (7C3 x 4C2)
 7  6 5 4 3
=


 3  2 1 2 1 
= 210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
=5x4x3x2x1
= 120
Required number of ways = (210 x 120) = 25200
5. In how many ways can the letters of the word 'LEADER' be arranged?
A) 72
B) 144
C) 360
D) 720
E) None of these
Answer: Option C
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways =
6!
 360
(1!)(2!)(1!)(1!)
6. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways
can they be selected such that at least one boy should be there?
A) 159
B) 194
C) 205
D) 209
E) None of these
Answer: Option D
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
 6 5 4 3  6 5 4
 65
= (6  4)  

 4  


 2 1 2 1   3  2 1
  2 1 
= (24 + 90 + 80 + 15)
= 209
7. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are
divisible by 5 and none of the digits is repeated?
A) 5
B) 10
C) 15
D) 20
Answer: Option D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1
way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5
ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of
filling it.
Required number of numbers = (1 x 5 x 4) = 20
8. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men
and 10 women?
A) 266
B) 5040
C) 11760
D) 86400
E) None of these
Answer: Option C
Explanation:
Required number of ways = (8C5 x 10C6)
= (8C3 x 10C4)
 8  7  6 10  9  8  7 
=


 3  2 1 4  3  2 1 
=11760.
9. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be
drawn from the box, if at least one black ball is to be included in the draw?
A) 32
Answer: Option C
Explanation:
B) 48
C) 64
D) 96
E) None of these
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
6 5  3 2


= 3 
 6  1

2 1   2 1


= (45 + 18 + 1)
= 64
10. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way
that the vowels occupy only the odd positions?
A) 32
B) 48
C) 36
D) 60
E) 120
Answer: Option C
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 3! = 6.
Total number of ways = (6 x 6) = 36.
11. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3
women?
A) 63
Answer: Option A
B) 90
C) 126
D) 45
E) 135
Explanation:
Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =
76 
 3   63

 2 1

12. How many 4-letter words with or without meaning, can be formed out of the letters of the
word, 'LOGARITHMS', if repetition of letters is not allowed?
A) 40
B) 400
C) 5040
D) 2520
Answer: Option C
Explanation:
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4
= (10 x 9 x 8 x 7)
= 5040
13. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so
that the vowels always come together?
A) 10080
B) 4989600
C) 120960
D) None of these
Answer: Option C
Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are
different.
Number of ways of arranging these letters =
8!
= 10080
(2! )(2! )
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =
4!
= 12.
2!
Required number of words = (10080 x 12) = 120960
14. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the
vowels always come together?
A) 120
B) 720
C) 4320
D) 2160
E) None of these
Answer: Option B
Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
15.How many alphabets need to be there in a language if one were to make 1 million distinct 3
digit initials using the alphabets of the language?
A) 26
B) 50
C) 100
D) 1000
Correct Answer - (C)
Solution:
1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.
Note distinct initials is different from initials where the digits are different.
For instance, AAA and BBB are acceptable combinations in the case of distinct initials while
they are not permitted when the digits of the initials need to be different.
This n3 different initials = 1 million
i.e. n3 = 106 (1 million = 106)
=> n3 = (102)3 => n = 102 = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.
FIFTY PROBLEMS WITH SOLUTIONS
1. In how many different ways may a class 1 spelunker who has never explored any of the
eightcaves before set up a tour of three caves, if she wishes to explore caves Abbott and
Caesar?
(a) 2
(b) 3
(c) 4
(d) 5
(e) 6
Ans. (b) Satyam Test Paper
2. Seven different toys are distributed among 3 children how many different ways are
possible?
(a)
7
C3
(b) 7P3
(c) 3 7
(d) 7 3
Ans. (c) Wipro
3. Find the total number of distinct vehicle numbers that can be formed using two
letters followed by two numbers. Letters need to be distinct. TCS
Ans:65000
This question comes under permutations and combinations section. Out of 26 alphabets
two distinct letters can be chosen in 26p2 ways. Coming to numbers part, there are 10 ways
(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10
ways to choose the second digit. Hence there are totally 10X10 = 100 ways. Combined with
letters there are 6p2 X 100 ways = 65000 ways to choose vehicle numbers.
4. If the letters of the word SACHIN are arranged in all possible ways and these
wordsare written out as in dictionary, then the word ‘SACHIN’ appears at serial
number
( a ) 601 ( b ) 600 ( c ) 603 ( d ) 602 [ AIEEE 2005 ]
Ans: a)
if the word started with the letter A then the remaining 5 positions can be filled in
5! Ways
If it started with c then the remaining 5 positions can be filled in 5! Ways
Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways
`
If it started with S then the remaining position can be filled with A,C,H,I,N
alphabetical order as on dictionary
in
The required word SACHIN can be obtained after the 5X5!=600 Ways
i.e. SACHIN is the 601th letter
5. How many ways are here to arrange the letters in the word GARDEN with the
vowelsin alphabetical order ?
( a ) 120 ( b ) 240 ( c ) 360 (d ) 480 [ AIEEE 2004 ]
Ans : c
6. If repetition of the digits is allowed, then the number of even natural numbers
havingthree digits is
( a ) 250 ( b ) 350 ( c ) 450 ( d ) 550 [ AIEEE 2002 ]
Ans: c)
In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)
10’s place can be filled in 10 different ways
100’s place can be filled in 9 different ways
There fore total number of ways = 5X10X9 = 450
7.
If n+1
C3 =2 nC2 , then the value of n is
(a)3
(b)4
(c)5
( d ) 6 [ AIEEE 2002 ]
Ans: (c)
8. The number of arrangements of the letters of the word BANANA in which the two
N’s do not appear adjacently is
( a ) 40
( b ) 60
( c ) 80
( d ) 100 [ IIT 2002 ]
Ans: (a)
9. Find the number of ways to arrange 4 people in groups of 3 at a time where order
matters?
A)20
B)24
C)24
D)30
Answer:(B)
Explanation:
There are 24 ways to arrange 4 items taken 3at a time when order matters.
10. Find the number of ways to arrange 6 items in groups of 4 at a time where order matters?
A)36
B)720
C)360
D)420
Answer:(C)
Explanation:
There are 360 ways to arrange 6 items taken 4 at a time when order matters.
11. Find the number of ways to take 4 people and place them in groups of 3 at a time where
order does not matter?
A)3
B)4
C)5
D)6
Answer:(B)
Explanation:
Since order does not matter, use the combination formula.
There are 4 ways to arrange 4 items taken 3 at a time when order does not matter.
12. Find the number of ways to take 20 objects and arrange them in groups of 5 at
where order does not matter.?
A)15504
B)40515
C)51540
D)15405
a time
Answer : (A)
Explanation:
There are 15,504 ways to arrange 20 objects taken 5 at a time when order does not matter
13. How many ways are there to select a subcommittee of 7 members from among a
committee of 17?
A)19000
B)19500
C)19448
D)19844
Answer : (C)
Explanation:
Since it does not matter what order the committee members are chosen in, the
combination formula is used.
Committees are always a combination unless the problem states that someone like a
president has higher hierarchy over another person. If the committee is ordered, then it
is a permutation.
C(17,7)= 19,448
14. Determine the total number of five-card hands that can be drawn from a deck of 52 cards.
A)2598960
B)2596800
C)2986800
D)2598708
Answer : ( A)
Explanation:
When a hand of cards is dealt, the order of the cards does not matter. If you are dealt
two kings, it does not matter if the two kings came with the first two cards or the last two
cards.
Thus cards are combinations. There are 52 cards in a deck and we want to
know how many different ways we can put them in groups of five at a time when order
does not matter. The combination formula is used.
C(52,5) = 2,598,960
15. Therefore there are 2,598,960 different ways to create a five-card hand from a deck of 52
cards. A school has scheduled three volleyball games, two soccer games, and four
basketball games. You have a ticket allowing you to attend three of the games. In how
many ways can you go to two basketball games and one of the other events?
A)6
B)5
C)30
D)11
Answer : (C)
Explanation :
Since order does not matter it is a combination.
The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.
We want 2 basketball games and 1 other event. There are 5 choices left.
C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball)*C(have 5 choices left, want 1)
C(4,2)*C(5,1)
(6)(5) = 30
Therefore there are 30 different ways in which you can go to two basketball games and one
of the other events.
16. How many ways are there to deal a five-card hand consisting of three eight's and two
sevens?
A)5
B)24
C)42
D)54
Answer : (B)
Explanation:
If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.
The hand still contains four Queens and an Ace.
problems with cards.
Thus, use the combination formula for
We have 4 eights and 4 sevens.
We want 3 eights and 2 sevens.
C(have 4 eights, want 3 eights)*C(have 4 sevens, want 2 sevens)
C(4,3)*C(4,2) = 24
Therefore there are 24 different ways in which to deal the desired hand.
17. Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and
ending with a 8?
A)1024
B)1042
C)4012
D)8!
Answer : (A)
Explanation:
There are 5 slots.
__ __ __ __ __ The first slot must be a four. There are 4 ways to put a four in the first slot.
There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third
slot. etc.
(4)(4)(4)(4)(4) = 45 = 1024
Therefore there are 1024 different ways to produce the desired hand of cards.
18. A certain marathon had 50 people running for first prize, second, and third prize.
A) How many different possible outcomes are there for the first three runners to cross the
finish line?
Solution:
Order matters in a race, so use the permutation formula.
P(50,3) = 117,600 ways
B) How many ways are there to correctly guess the first, second, and third place winners?
Solution:
There is 1 way to correctly guess who comes in first, second, and third. There is only one
set of first, second and third place winners. You must correctly guess these three people,
and there is only one way to do so.
19. A local delivery company has three packages to deliver to three different homes.
if the packages are delivered at random to the three houses, how many ways are there
for at least one house to get the wrong package?
A)3
B)5
C)3!
D)5!
Answer : (B)
Solution:
The possible outcomes that satisfy the condition of "at least one house gets the wrong
package" are:
One house gets the wrong package or two houses get the wrong package or three houses get
the wrong package.
We can calculate each of these cases and then add them together, or approach this problem
from a different angle.
The only case which is left out of the condition is the case where no wrong packages are
delivered.
If we determine the total number of ways the three packages can be delivered and then
subtract the one case from it, the remainer will be the three cases above.
There is only one way for no wrong packages delivered to occur. This is the same as
everyone gets the right package. The first person must get the correct package and the second
person must get the correct package and the third person must get the correct package.
1*1*1 = 1
Determine the total number of ways the three packages can be delivered.
3*2*1 = 6
The number of ways at least one house gets the wrong package is:
6-1=5
Therefore there are 5 ways for at least one house to get the wrong package.
20. Write down all the permutations of xyz.
Sol:xyz, xzy, yxz, yzx, zxy, zyx.
21. How many permutations are there of the letters pqrs?
A)4
B)4^2
C)2^4
D)8
Sol:4! = 1· 2· 3· 4 = 24
22. .How many different four letter words can be formed (the words need not be meaningful)
using the letters of the word "MEDITERRANEAN" such that the first letter is E and the
last letter is R?
A).59
B)
11!
2!*2!*2!
C.56
D.23
E.
11!
3!*2!*2!*2!
The correct choice is (A) and the correct answer is 59
Explanatory Answer
The first letter is E and the last one is R.
Therefore, one has to find two more letters from the remaining 11 letters.
Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.
The second and third positions can either have two different letters or have both the
letters to be the same.
Case 1: When the two letters are different. One has to choose two different letters from
the 8 available different choices. This can be done in 8 * 7 = 56 ways.
Case 2: When the two letters are same. There are 3 options - the three can be either Ns or
Es or As. Therefore, 3 ways.
Total number of possibilities = 56 + 3 = 59
23. In how many ways can the letters of the word ABACUS be rearranged such that the
vowels always appear together?
3!*3!
4!*3!
E.
2!
2!
4!*3!
The correct choice is (D) and the correct answer is
.
2!
A.
6!
2!
B)3!*3!
C)
4!
2!
D.
Explanatory Answer
ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3
vowels together.
These 4 elements can be rearranged in 4! Ways.
The 3 vowels can rearrange amongst themselves in
3!
ways as "a" appears twice.
2!
Hence, the total number of rearrangements in which the vowels appear together are
4!*3!
2!
24. .In how many ways can the letters of the word "PROBLEM" be rearranged to make 7
letter words such that none of the letters repeat?
B)7C7
A. 7!
C)77
D)49
The correct choice is (A) and the correct answer is 7!.
Explanatory Answer
There are seven positions to be filled.
The first position can be filled using any of the 7 letters contained in PROBLEM.
The second position can be filled by the remaining 6 letters as the letters should not repeat.
The third position can be filled by the remaining 5 letters only and so on.
Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7!
Ways.
25. How many lines can you draw using 3 non collinear (not in a single line) points A, B and
C on a plane?
A) 6
B)
4
C)
3
D)2
Solution:
You need two points to draw a line. The order is not important. Line AB is the same as line
BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did
with permutations, we get the following pairs of points to draw lines.
AB , AC
BA , BC
CA , CB
There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and
CB.
The lines are: AB, BC and AC ; 3 lines only.
So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the
points A, B and C is not important.
26. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can
take more than 10 letters?
(A)5^10
(B)10^5
Correct Answer - (A)
(C)10P5
(D) 10C5
Solution:
Each of the 10 letters can be posted in any of the 5 boxes.
So, the first letter has 5 options, so does the second letter and so on and so forth for all of the
10 letters.
i.e. 5*5*5*….*5 (upto 10 times)
= 5 ^ 10.
27. How many number of times will the digit ‘7' be written when listing the integers from 1
to 1000?
(A)271
(B)300
(C)252
(D) 304
Correct Answer - (B)
Solution:
7 does not occur in 1000. So we have to count the number of times it appears between 1
and 999. Any number between 1 and 999 can be expressed in the form of xyz where
0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other
9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second
or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and
3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with
the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or
m second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
243 + 54 + 3 = 300
28. . A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the
other only 4. In how many ways can they travel?
(A)9
(B)26
(C)126
(D) 3920
Correct Answer - (C)
Solution:
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows
Case I: 5 students in the first car and 3 in the second
Or Case II: 4 students in the first car and 4 in the second
Hence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.
Therefore, the total number of ways in which 8 students can travel is
8
C3 + 8C4 = 56 + 70 = 126.
29. What is the value of 1*1! + 2*2! + 3*3! + ............ n*n!,
where n! means n factorial or n(n-1)(n-2)...1
(A)n(n-1)(n-1)! (B)(n+1)!/(n(n-1))
Correct Answer - (D)
Solution:
(C)(n+1)!-n!
(D) (n + 1)! - 1!
1*1! = (2 -1)*1! = 2*1! - 1*1! = 2! - 1!
2*2! = (3 - 1)*2! = 3*2! - 2! = 3! - 2!
3*3! = (4 - 1)*3! = 4*3! - 3! = 4! - 3!
..
..
..
n*n! = (n+1 - 1)*n! = (n+1)(n!) - n! = (n+1)! - n!
Summing up all these terms, we get (n+1)! - 1!
30. Eight first class and six second class petty officers are on the board of the 56 club. In
how many ways can the members elect, from the board, a president, a vice-president,
a secretary, and a treasurer if the president and secretary must be first class petty officers
and the vice-president and treasurer must be second class petty officers?
SOLUTION: Since two of the eight first class petty officers are to fill two different
offices, we write
8 P2 
8!
8!

 8 . 7  56
( 8  2 )!
6!
Then, two of the six second class petty officers are to fill two different offices; thus, we write
6 P2 
6!
6!

 6 . 5  30
( 6  2 )!
4!
The principle of choice holds in this case; therefore, the members have
56.30 = 1,680ways to select the required office holders
31. A standard deck of playing cards has 13 spades. How many ways can these 13 spades be
arranged?
A)13x13
B) 13
C)
13!
D) (13+13)!
E)None of these
Solution:
The solution to this problem involves calculating a factorial. Since we want to know how 13
cards can be arranged, we need to compute the value for 13 factorial.
13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800
32. A coach must choose five starters from a team of 12 players. How many different ways can
the coach choose the starters?
A) 789 B)
987
C)
791 D) 792 E)297
Answer:(D)
Solution:
Choose 5 starters from a team of 12 players. Order is not important.
33. There are fourteen juniors and twenty-three seniors in the Service Club. The club is to send
four representatives to the State Conference. If the members of the club decide to send two
juniors and two seniors, how many different groupings are possible?
A) 23000 B) 22032 C)23023 D) 22022 E) None of these
Answer : (C)
Solution:
Choose 2 juniors and 2 seniors.
34.There are 2 brothers among a group of 20 persons. In how many ways can the group be
arranged around a circle so that there is exactly one person between the two brothers?
(A)2*17!
(B)18!*18
(C)19!*18
(D)2*18!
(E)2*17!*17!
Correct Answer is 2 * 18! - Choice (D)
Explanation
Circular Permutation 'n' objects can be arranged around a circle in (n - 1)!.
If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the
number arrangements will be half that number.
(n  1)!
i.e., number of arrangements =
.
2
Let there be exactly one person between the two brothers as stated in the question.
If we consider the two brothers and the person in between the brothers as a block, then there will
17 others and this block of three people to be arranged around a circle.
The number of ways of arranging 18 objects around a circle is in 17! ways.
Now the brothers can be arranged on either side of the person who is in between the brothers in
2! ways.
The person who sits in between the two brothers could be any of the 18 in the group and can be
selected in 18 ways.
Therefore, the total number of ways 18 * 17! * 2 = 2 * 18!.
35. How many integers, greater than 999 but not greater than 4000, can be formed with the digits
0, 1, 2, 3 and 4, if repetition of digits is allowed?
(A)499
(B)500
(C)375
(D)376
(E)501
Correct Answer is 376 - Choice (D)
Explanation
The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one
of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1
or 2 or 3 or 4.
Hence, there are 3 * 5 * 5 * 5 or 375 numbers from 1000 to 3999.
Including 4000, there will be 376 such numbers.
36. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is
empty, if any of the boxes may hold all of the toys?
(A)20(B)30(C)25(D)600(E)480
Correct Answer is 25 - Choice (C)
Explanation:
The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2, 2, 1
b. 3, 1, 1
Case a. Number of ways of achieving the first option 2 - 2 - 1
Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be
selected in 3C2 ways and the last toy can be selected in 1C1 way.
However, as the boxes are identical, the two different ways of selecting which box holds the first
two toys and which one holds the second set of two toys will look the same. Hence, we need to
divide the result by 2.
Therefore, total number of ways of achieving the 2 - 2 - 1 option is
5C 2 * 3C 2 10 * 3

 15 ways
2
2
Case b. Number of ways of achieving the second option 3 - 1 - 1
Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining
two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case a + number of ways of achieving Case b
= 15 + 10 = 25 ways.
37 . A key pad lock has 10 different digits, and a sequence of 5 different digits must be selected
for the lock to open. How many key pad combinations are possible?
A)24030
B) 30240
C) 30201
D) 12001 E)None of these
Solution:(B)
First we need to find n and r :
n is the number of digits we have to choose from. If n is 10 then there are 10 digits.
r is the number of digits being used at a time. If r is 5,then there are 5 digits in the sequence.
Putting this into the permutation formula we get:
38 . How many necklaceof 12 beads each can be made from 18 beads of different colours?
A) B)
C) D)
E)
Ans. Here clock-wise and anti-clockwise arrangements are same.
18
Hence total number of circular–permutations:
P12/2x12 = 18!/(6 x 24)
39. In how many ways, can zero or more letters be selected form the letters AAAAA?
A)
B)
C) D)
E)
Ans. Number of ways of :
A) Selectingzero'A's= 1
Selecting one 'A's = 1
Selecting two 'A's = 1
Selecting three 'A's = 1
Selecting four 'A's = 1
Selecting five 'A's = 1
=> Required number ofways =6
[5+1]
40. In how many ways can the letters of the word 'MISSISIPPI' be arranged?
a)1260b)12000c)12600d)14800 e) 26800
Answer: (C )
soution:
Total # of alphabets = 10
so ways to arrange them = 10!
Then there will be duplicates because 1st S is no different than 2nd S.
we have 4 Is 3 S and 2 Ps
Hence # of arrangements = 10!/4!*3!*2!
41. Goldenrod and No Hope are in a horse race with 6 contestants. How many different
arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of
the horses finish the race?
(A) 720(B) 360(C) 120(D) 24(E) 21
Answer (B)
Solution:
two horses A and B, in a race of 6 horses... A has to finish before B
if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish
in 4! Ways, so total ways 5*4!
if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other
positions could be filled in 4! ways, so the total ways 4*4!
if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions
could be filled in 4! ways, so total ways 3*4!
if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions
could be filled in 4! ways, so total ways... 2 * 4!
if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..
A cannot finish 6th, since he has to be ahead of B
therefore total number of ways
5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360
42. If 5 nP3 = 4 n+1P3 find n?
A)10
B) 12
C) 11
D)14
Answer:(D)
Solution:
nP3 = n(n – 1) (n – 2)
n+1P3 = (n + 1) n (n – 1) [Replacing n by (n + 1)]
∴ 5 n(n −1)(n − 2) = 4(n +1)(n)(n −1)
Divide both sides by n (n-1)
5(n − 2) = 4(n +1)
5n −10 = 4n + 4
5n − 4n = 4 +10
n = 14
43. There are 7 non-collinear points. How many triangles can be drawn by joining these
points?
A) 35
B)10
C)8
D)7
Solution: A,
A triangle is formed by joining any three non-collinear points in pairs.
There are 7 non-collinear points
The number of triangles formed = 7C3
 7  (7  1)  (7  2) 


=
3!

7 65
=

 3  2 1 
=7x5
= 35
44. How many diagonals can be drawn in a pentagon?
B) 5
B)10
C)8
D)7
Answer:(A)
Solution. A pentagon has 5 sides. We obtain the
diagonals by joining the vertices in pairs.
Total number of sides and diagonals
= 5C2
5 4
=

 2 1 
= 5 x 2 = 10
This includes its 5 sides also.
Diagonals = 10 – 5 = 5
Hence the number of diagonals = 10 – 5 = 5
45.From 8 gentlemen and 4 ladies, a committee of 6 is to be formed. In how many
ways can this be done so that the committee contains exactly 2 ladies?
A)6
B)70
C) 76
D)420
Answer:(D)
Explanation: Exactly two ladies.
Ladies
4
2
But
Gentlemen
8
4
Number of ways
4C2 x 8C4
4C2 = 6 and 8C4 = 70
The number of ways = 6 x 70 = 420
46) There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can
a team of 11 be selected so that the team contains at least 4 bowlers?
A)126
B)504
C)540
D)1170
Answer:(D)
Explanation:
Possibilities
1
2
3
Bowlers Batsmen
6
9
4
7
5
6
6
5
6C4 x 9C7 = 15 x 36 = 540
6C5 x 9C6 = 6 x 84 = 504
6C6 x 9C5 = 1 x 126 = 126
Total = 1170
47. If 6Pr = 360 and 6Cr = 15 find r ?
A)4
Answer:(A)
B)3
Explanation: nPr = nCr x r !
6Pr = 15 x r!
360 = 15 x r!
r! = 360/15
= 24
=4x3x2x1
r! = 4!
Therefore r = 4
C)6
D)5
Number of ways
6C4 x 9C7
6C5 x 9C6
6C6 x 9C5
QUESTION BANK
1. There are five women and six men in a group. From this group a committee of 4 is to be
chosen. How many different ways can a committee be formed that contain three women
and one man?
A)55
B)60
C)25
D)192
[
]
2. There are five women and six men in a group. From this group a committee of 4 is to be
chosen. How many different ways can a committee be formed that contain at least three
women?
A)65
B)67
C)20
D)12
[
]
[
]
3. How many different 5-card hands include 4 aces?
A)61
B)28
C)48
D)225
4. In a local election, there are seven people running for three positions. The person that
has the most votes will be elected to the highest paying position. The person with the
second most votes will be elected to the second highest paying position, and likewise for
the third place winner. How many different outcomes can this election have?
A)301
B)258
C)176
D)210
[
]
5. How many different ways can a five-question true-false test can be answered?
A)32
B)28
C)76
D)26
[
]
6. How many ways can 10 people be placed in alphabetical order according to their first
names?
A)1
B)10
C)100
D)10!
[
]
7. How many different ways can the letters in the word "STORE" be arranged?
A)15
B)5
C)120
D)24
[
]
8. How many different ways are there to arrange the word “SCHOOL”?
A)455
B)53
C)487
D)360
[
]
9. How many words with or without meaning, can be formed by using all letters of the
word, “DELHI”, using each letter exactly once?
A)10
B)25
C)60
D)120
[
]
10. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be
formed?
A)210
B)1050
C)25200
D)21400
[
]
11. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8
men and 10 women?
A)2660
B)5040
C)11760
D)86400
[
]
[
]
12. In how many ways 7 different gems can be arranged in a necklace?
A)120
B)240
C)360
D)480
13. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls
be drawn from the box, if atleast one black ball is to be included in the draw?
A)32
B)48
C)64
D)96
[
]
14. How many 3-digit numbers are possible using the digits 0, 1, 4,5, 7 and 9?
A) 180
B) 120
C)100
D) 140
[
]
[
]
15. Find the number of combinations of 7 objects taking 2 at a time?
A) 12
B) 10
C)
21
D) 14
16. A class has 14 boys and 16 girls. The class has to elect one boy and one girl to be
representatives on a committee. How many different ways can they select them?
A) 114
B) 104
C) 224
D) 212
[
]
17. How many arrangements of three types of flowers are there if there are 6 types to choose
from?
A) 12
B) 18
C)20
D)
16
[
]
18. If a university student has to choose 2 science classes from 5 available science classes
and 3 other classes from a total of 7 other classes available, how many different groups of
classes are there?
A) 225
B)
350
C) 175
D) 348
[
]
19. How many ways can 7 people be arranged around a roundtable?
A) 270
B)
720
C) 340
D) 625
[
]
20. A license plate has 3 letters and 3 digits in that order. A witnessto a hit and run accident
saw the first 2 letters and the last digit. If the letters and digits can be repeated, how many
license platesmust be checked by the police to find the culprit?
A) 1300
B) 1800
C) 2600
D) 2200
[
]
21. How many different ways can you have 1st, 2nd, 3rd, and 4thin a race with 10 runners?
A) 5040
B) 2800
C) 3684
D) 4840
[
]
22. You select a president and vice-president from a group of 5students. Find the number of
possible outcomes?
A)
18
B) 20
C) 28
D)24
[
]
23. You must select a committee of 3 from 12 students. How manydifferent committees can
be formed?
A) 259
B) 360
C)
480
D) 720
[
]
24. How many triangles can you make using 6 non collinear points on a plane?
A)
6
C1=6
6
B)
C4=15
C)
6
C6=1
D)
6
C3=20
[
]
25. In how many ways can you arrange 5 different books on a shelf?
A) 5*5=25
B)
26. Find the value of
A)
1
5!=120
C) 4!=24
D) 30
[
]
[
]
7 P6  7 P5
.
7 P6  7 P5
B )2
C) 3
D) 6
27. How many different ways can 4 red, 3 blue, 4 yellow, and 2 green bulbs be arranged on
a string of Christmas tree lights with 13 sockets?
A)
900
B)
90
C) 90,90
D)
900,900
[
]
28. On each trip, a salesman visits 4 of the 12 cities in his territory. In how many different
ways can he schedule his route?
A)225
B)495
C)285
D) 197
[
]
29. How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9
inclusive?
A)84
B)120
C)504
D)720
[
]
30. How many permutations of 4 different letters are there, chosen from the twenty six letters
of the alphabet?
A) 14,950
B) 23,751
C)358,800
D) 456,976
[
]
31. A password consists of four different letters of the alphabet. How many different possible
passwords are there ?
A) 426
B)456,976
C)14,950
D)358,800
[
]
32. Restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 choices of
dessert. A customer can choose to eat just one course, or two different courses, or all
three courses. Assuming all choices are available, how many different possible meals
does the restaurant offer?
A) 329
B)310
C)200
D)19
[
]
33. In a contest in which there are 8 participants, in how many ways can 5 distinct prizes be
Awarded?
A.) 112
B.) 6720
C.) 336
D.) 672
[ ]
34. A club elects a president, vice-president, and secretary-treasurer. How many sets of
officers are possible if there are 15 members and any member can be elected to each
position? No person can hold more than one position.
A.) 2730
B.) 32,760
C.) 910
D.) 1365
[ ]
35. A church has 7 bells in its bell tower. Before each church service 5 bells are rung in
sequence. No bell is rung more than once. How many possible sequences are there?
A.) 2520
B.) 42
C.) 84
D) 21
[
]
36. How many arrangements can be made using 2 letters of the word HYPERBOLAS if no
letter is to be used more than once?
A.) 1,814,400 B.) 3,628,800 C.) 45
D.) 90
[ ]
37. A work softball team has 15 players on its roster. There are 9 distinct positions in which
these players can be placed. How many lineups can be fielded?
A.) 1,505,667,870 B.) 1,635,890 C.) 1,816,214,400
D.) 214,400 [ ]
38. From a group of 8 people, 5 will each win $1,000. How many different winning groups
are possible?
A.) 56
B.) 6720
C.) 168
D.) 336
[
]
39. Of a classroom filled with 20 students, 2 will be selected to stay after school and correct
homework for extra credit. How many combinations are possible?
A.) 190
B.) 210
C.) 63
D.) 40
[
]
40. To win the lottery, one must correctly select 6 numbers from a collection of 50 numbers
(one through 50). The order in which the selection is made does not matter. How many
different selections are possible?
A.) 250
B.) 15,890,700
C.) 300
D.) 13,983,816
[
]
41. A test is administered with 15 questions. Students are allowed to answer any ten. How
many choices of ten questions are there?
A.) 150
B.) 250
C.) 3003
D.) 3000
[
]
42. There are 20 people who work in an office together. Four of these people are selected to
go to the same conference together. How many such selections are possible?
A.) 116280
B.) 4845
C.) 3003
D.) 80
[
]
43. There are 20 people who work in an office together. Four of these people are selected to
attend four different conferences. The first person selected will go to a conference in New
York, the second will go to Chicago, the third to San Franciso, and the fourth to Miami.
How many such selections are possible?
A.) 116280
B.) 80
C.) 3003
D.) 4845
[
]
44. Serial numbers for a product are to be made using three letters (using any letter of the
alphabet) followed by two single-digit numbers. For example, JGR29 is one such serial
number. How many such serial numbers are possible if neither letters nor numbers can be
repeated
A.) 1,404,000 B.) 80
C.) 15690
D.) 117,000
[
]
45. A 7-card hand is chosen from a standard 52-card deck. How many of these will have four
spades and three hearts
A.) 1001
B.) 204,490
C.)29,446,560
D.) 117,000
[
]
46. In a new group of 15 employees at a restaurant, 10 are to be assigned as servers, 3 are to
be assigned as hosts, and 2 are to be assigned as cashiers. In how many ways can the
assignment be made?
A.) 60
B.) 204,490
C.)30,030
D.) 3014
[
]
47. In how many ways can a first prize, a second prize and four identical third prizes be
awarded to a group of 15 people?
A.) 60
B.) 150,150
C.)5005
D.) 3,603,600
[
]
48. There are 10 students from whom 4 are going to be chosen to represent their school at a
conference. If Jack, Anna or Chris, but only one of them, must be chosen, in how many
ways can the students be chosen to go to the conference ?
A.) 210
B.) 150,150
C.)360
D.) 105
[
]
49. There are 20 people who work in an office together. Four of these people are selected to
attend four different conferences. The first person selected will go to a conference in New
York, the second will go to Chicago, the third to San Franciso, and the fourth to Miami.
How many such selections are possible ?
A.) 116,280
B.) 80
C.)4845
D.) 105
[
]
50. If the letters of the word SACHIN are arranged in all possible ways and these words
are written out as in dictionary, then the word ‘SACHIN’ appears at serial number
( A ) 601
(B)
600
( C ) 603
( D ) 602
[
]
ANSWERS
1
B
21
A
41
C
2
A
22
B
42
B
3
C
23
D
43
A
4
D
24
D
44
A
5
A
25
B
45
B
6
A
26
C
46
C
7
C
27
D
47
B
8
D
28
B
48
D
9
D
29
D
49
A
10
C
30
C
50
A
11
C
31
D
12
C
32
C
13
C
33
B
14
A
34
A
15
C
35
A
16
C
36
D
17
C
37
C
18
B
38
A
19
B
39
A
20
C
40
B