COGNIZANT Permutation and Combinations Questions with Solutions

  1. The number of ways in which 15 students A1,A2,—A15 can be ranked, such that A4 is always above A8 is:

a)15!
b)13!
c)15!/2
d)13!/2
Answer:
Explanation: When A8 is in 2nd position A4 can occupy only 1st. When A8 is in 3rd position A4 can occupy only 1st or 2nd.when A8 in 4th A4 can occupy 1st or 2nd or 3rd. Number of ways in which A4 and A8 can be arranged = 1 + 2 +3+ …+14 = 14 * 15 In the remaining 13 positions other students can be permuted in 13! ways. Hence total number of ways = 13!/2 * 14 * 15= 15! /

2.  From a pack of 52 playing cards, 4 cards are removed at random. In how many ways can the 1st place and 3rd place cards be drawn out such that both are black ?

A: 64,974
B: 62,252
C: 69,447
D:1,592,500
Answer:
Explanation: There are 26 black cards to choose from on the first draw.
Now the second card can either be black or red.
If black, then there are 25 to choose from, and then for
the third card there will then be 24 black cards to choose
from.If red, then there are 26 reds to choose from, and then
for the third card there are then 25 black cards tochoose from.
In either case, there are 49 cards left to from on the
4th draw.
So the number of ways we can do this is
26*(25*24 + 26*25)*49 = 1,592,500

3 . 20 Men can do a job in 10 days, working 8 hrs a day if women is 33.33% more efficient than men. How many women will it take to finish the same job in 10days working 6 hrs a day?

A) 10
B) 12
C) 15
D) 16
E) 20
Answer:E
Explanation: By using formula:( M1D1H1/W1)=(M2D2H2/W2)
===>(20*10*8/133.33)=(M2*10*6/100)
===>M2=20

4.There are n different books and p copies of each. the number of ways in which a selection can be made from them is

A: np
B: pn
C: (p+1)n -1
D: (n+1)p-1
Answer:C
Explanation: We have ‘p’ copies of each of ‘n’ different books.
Number of ways to select from p copies:
p + 1
Total no. of ways:
(p + 1)(p + 1))p + 1)…………….n times
(p + 1)^n
Number of different ways of selection is
[(p + 1)^n]– 1

5. In how many ways can 4 ladies and 4 men form two mixed doubles teams for a tennis match?

A: 72
B: 108
C: 36
D: 84
Answer:A
Explanation: two mixed double pair can be obtained in 4C2 * 4C2 ways which is 36
and this two pair can be ordered in 2 ways i.e 2 * ( 4C2 * 4C2 ) which is 72

6. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made

A)1260
B)1550
C)1472
D)none
Answer:A
Explanation: A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made

7. How many 5 digit no. can be formed from 0,2,4,5 & 9 ?

A)95
B)96
C)120
D)125
Answer:B
Explanation: 5!-4!
We can arrange those 5 letters in 5! ways and If 0 comes at the ten thousands place it will not be an 5 digit number. So the possibility of getting 0 at the 1st place is 4! ways. So we need to subtract this 4! from the 5! to get the answer.

  1. 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: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.

  1. 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.

  1. 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: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.

  1. A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 women. The shortlist consists of 9 men and 6 women. In how many ways can this be done?

A: 3,724
B: 3,630
C: 4,914
D: 5,670
Correct Answer : D
Explanation: Three possibilities: (1W+6M), (2W+5M), (3W+4M)
=> (6C1 *9C6)+(6C2 *9C5)+(6C3 *9C4)= 4914.

  1. From a pack of 52 playing cards, 4 cards are removed at random. In how many ways can the 1st place and 3rd place cards be drawn out such that both are black ?

A: 64,974
B: 62,252
C: 69,447
D: 1,592,500
Correct Answer : D
Explanation: There are 26 black cards to choose from on the first draw.
Now the second card can either be black or red.
If black, then there are 25 to choose from, and then for
the third card there will then be 24 black cards to choose
from.If red, then there are 26 reds to choose from, and then
for the third card there are then 25 black cards to
choose from.
In either case, there are 49 cards left to from on the
4th draw.
So the number of ways we can do this is
26*(25*24 + 26*25)*49 = 1,592,500,

  1. In how many ways can 4 ladies and 4 men form two mixed doubles teams for a tennis match?

A)65
B)72
C)85
D)none
Answer:B
Explanation: 2 women and 2 men in
4c2*×4c2 = 36 ways
and pair them in 2 ways to get ans = 72

  1. A boy has 4 different boxes and 5 different marbles. In how many ways can he place the marbles in the boxes such that each box has at least one marble ?

A)488
B)480
C)120
D)963
Answer:C
Explanation: Let the boxes be b 1, b 2, b 3 and b 4
so,
box 1 can be filled in ways = 5
box 2 can be filled in ways = 4
box 3 can be filled in ways = 3
box 4 can be filled in ways = 2
so, 5*4*3*2 = 120 ways
Lastly any of 4 can be filled with 5th marble in 4 ways.
So, 120*4 = 480 ways

  1. A box contains 20 tickets of identical appearance, the tickets being numbered 1, 2, 3, ….., 20. In how many ways can 3 tickets be chosen such that the numbers on the drawn tickets are in arithmetic progression ?

A: 18
B: 33
C: 56
D: 90
Correct Answer : D
Explanation: there could be maximum Common difference will be 9 so we have to check no of ways of AP with Common difference form 1 to 9
with C.D. 1 : 18 ways
with C.D. 1 : 16 ways
with C.D. 1 : 14 ways
with C.D. 1 : 12 ways
with C.D. 1 : 10 ways
with C.D. 1 : 8 ways
with C.D. 1 : 6 ways
with C.D. 1 : 4 ways
with C.D. 1 : 2 ways
TOtal No. of ways are 2 + 4 + 6 + 8 + 10 + 12 + 14 +16 + 18
with is 90

16. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

A)6!/2!
B) 3!*3!
C) (3!*3!)/2!
D)4!*3!)/2!
Answer:D
Explanation: ABACUS–> (vowels)consonants: (AAU)BCS => total 4 characters. 1-> AAU and 3-> B,C,S.
so arrangement of this is = 4 !
now vowels having it’s own arrangements= 3! but repetition of A is twice in vowels.
so actual arrangement for vowel= 3!/2!
So ans = 4!*3!/2!

17.  In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

A. 47200
B. 48000
C. 42000
D. 50400
Answer:
Explanation: The word ‘CORPORATION’ has 11 letters. It has the vowels ‘O’,’O’,’A’,’I’,’O’ in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO).
Hence we can assume total letters as 7. But in these 7 letters, ‘R’ occurs 2 times and rest of the letters are different.
Number of ways to arrange these letters
=7!/2!=(7×6×5×4×3×2×1)2×1=2520
In the 5 vowels (OOAIO), ‘O’ occurs 3 and rest of the vowels are different.
Number of ways to arrange these vowels among themselves =(5!/3!=5×4×3×2×1)/(3×2×1)=20
Hence, required number of ways
=2520×20=50400

  1. In how many ways a committee, consisting of 4 men and 10 women can be formed from 6men and 10 women?

A: 266
B: 50
C: 15
D: 8640
Correct Answer : C
Explanation:

Required number of ways = (8C5 x 10C6)
= (8C3 x 10C4)
8 x 7 x 6 x 10 x 9 x 8 x 7
3 x 2 x 1 4 x 3 x 2 x 1
= 11760.
  1. Out of 7 consonants and four vowels ,how many words of three consonants and 2 vowels can be

formed?
A)210
B)1050
C)25200
D)21440
Answer:C
Explanation: Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)
= 7 x 6 x 5 x 4 x 3
3 x 2 x 1 2 x 1
= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.

Number of ways of arranging5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.

Required number of ways = (210 x 120) = 25200.

20. The number of arrangements that can be made with the letters of the word MEADOWS so that the vowels occupy the even places?

A. 720
B. 144
C. 120
D. 36
Answer: Option B
Explanation:
The word MEADOWS has 7 letters of which 3 are vowels.
-V-V-V-
As the vowels have to occupy even places, they can be arranged in the 3 even places in 3! i.e., 6 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 6 = 14