SNAP Permutation and Combination Questions PDF

Permutation and Combination is an important topic in the Quant section of the SNAP Exam. You can also download this Free Permutation and Combination Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Permutation and Combination questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

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Question 1: In how many different ways can the letters of the word ‘STRESS’ be arranged

a) 360

b) 240

c) 720

d) 120

e) None of these

1) Answer (D)

Solution:

Total number of letters = 6 of which 3 are the same.

Therefore, number of ways of arranging = $\frac{6!}{3!}$ = $\frac{720}{6}$ = 120

Question 2: A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow?

a) 5/33

b) 5/22

c) 7/22

d) 7/33

e) 7/66

2) Answer (C)

Solution:

None yellow => both blue => is possible in $^7C_2$ ways

=> probability = $\frac{^7C_2}{^{12}C_2}$ = $\frac{7}{22}$

Question 3: A die is thrown twice. What is the probability of getting a sum 7 from the two throws?

a) 5/18

b) 1/18

c) 1/9

d) 1/6

e) 5/36

3) Answer (D)

Solution:

Number of ways in which the sum is  7 = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) => 6 ways

Total number of possibilities = 6 * 6 = 36

Probability that sum is 7 = $\frac{6}{36}$ = $\frac{1}{6}$

Question 4: In how many different ways can the letters of the word ‘THERAPY’ be arranged so that the vowels never come together?

a) 720

b) 1440

c) 5040

d) 3600

e) 4800

4) Answer (D)

Solution:

Number of ways of arranging seven letters = 7!
Let us consider the two vowels as a group
Now the remaining five letters and the group of two vowels = 6
These six letters can be arranged in 6!2! ways( 2! is the number of ways the two vowels can be arranged among themselves)
The number of ways of arranging seven letters such that no two vowels come together
= Number of ways of arranging seven letters – Number of ways of arranging the letters with the two vowels being together
= 7! – (6!2!)
= 3600

Question 5: A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?

a) $\frac{41}{190}$

b) $\frac{21}{190}$

c) $\frac{59}{290}$

d) $\frac{99}{190}$

e) $\frac{77}{190}$

5) Answer (D)

Solution:

Probability of getting two white balls = ¹³P₂
Probability of getting two black balls = ⁷P₂
Total probability = (¹³P₂ + ⁷P₂)/²⁰P₂ = 198/380 = 99/190

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Question 6: If 6 boys and 6 girls have to sit in a round circular music chair. So, that there is a girl between every 2 boys. Find the number of ways they can sit?

a) 6! × 5!

b) 6! × 4!

c) 6! × 3!

d) 6! × 2!

e) None of these

6) Answer (A)

Solution:

Circular permutation = n! (n – 1)!

∴ Number of ways = 6! (6 – 1)! = 6! × 5!

Question 7: What is the number of words formed from the letters of the word ‘JOKE’ So that the vowels and consonants alternate?

a) 4

b) 8

c) 12

d) 18

e) None of these

7) Answer (B)

Solution:

Word name: ‘JOKE’ Vowels: O, E
Consonants: J, K
∴ Possible arrangement beginning with consonant: JOKE, KOJE, JEKO, KEJO = 4 Numbers
beginning with vowel: OJEK, OKEJ, EJOK, EKOJ = 4 Numbers
Required number = 4+4 = 8 numbers

Question 8: In how many different ways can 5 men and 3 women be seated in a row such that no two women are next to each other?

a) 12200

b) 14400

c) 15600

d) 16400

e) None of the above

8) Answer (B)

Solution:

No two women would be seated next to each other if they sit between men. So, first the men can be arranged in 5! ways. There are six spots between the men. Number of ways to arrange the 3 women in those six spots is $ P_6^3 $ ways. So, total number of ways = 120*120 = 14400 ways.

Question 9: In how many different ways can the letter of the word ‘SIMPLE’ be arranged ?

a) 520

b) 120

c) 5040

d) 270

e) None of these

9) Answer (E)

Solution:

If there are n different letter in a word then n! different words can be formed.

In SIMPLE there are 6 different letters and hence 6! new words can be formed by different arangement.

Hence, 720 new words can be formed.

Question 10: In how many different ways can the letters of the word ‘SECOND’ be arranged ?

a) 720

b) 120

c) 5040

d) 270

e) None of these

10) Answer (A)

Solution:

If a word has n different letters in it, then they can be arranged in n! different ways.

Here SECOND has 6 different letters and they can be arranged in 6! ways i.e 720 different ways.

Instructions

Study the given information carefully and answer the question that follow:
A committee of five members is to be formed out of 3 trainees 4 professors and 6 research associates In how many different ways can this be done if____

Question 11: The committee should have 2 trainees and 3 research associates ?

a) 15

b) 45

c) 60

d) 9

e) None of these

11) Answer (C)

Solution:

We have to select 2 trainees out of 3 trainees and 3 resarch associates out of 6 resarch associates.

Required number of ways = $^3C_2*^6C_3 = 60$ ways

Question 12: The committee should have all 4 professors and 1 research associate or all 3 trainees and 2 professors ?

a) 12

b) 13

c) 24

d) 52

e) None of these

12) Answer (A)

Solution:

4 professors and 1 resarch associate can be selected in $^4C_4*^6C_1 = 6$ ways

Similarly 3 trainees and 2 professors would be selected in $^3C_3*^4C_2$ ways = 6

Total ways = 6+6 = 12

Instructions

Study the given information carefully and answer the question that follow:
A basket contains 4 red 5 blue and 3 green marbles.

Question 13: If the three marbles are picked at random what is the probability that at least one is blue ?

a) $\frac{7}{12}$

b) $\frac{37}{44}$

c) $\frac{5}{12}$

d) $\frac{7}{44}$

e) None of these

13) Answer (B)

Solution:

Required probability = 1 – Prob. that no ball is blue = $1- \frac{^7C_3}{^{12}C_3} = 37/44$

Question 14: If two marbles are drawn at random what is the probability that both are red ?

a) $\frac{3}{7}$

b) $\frac{1}{2}$

c) $\frac{2}{11}$

d) $\frac{1}{6}$

e) None of these

14) Answer (E)

Solution:

Probability that both the balls are red = $\frac{^4C_2}{^{12}C_2} = \frac{1}{11}$

Question 15: If three marbles are picked at random what is the probability that either all are green or all are red ?

a) $\frac{7}{44}$

b) $\frac{7}{12}$

c) $\frac{5}{12}$

d) $\frac{1}{44}$

e) None of these

15) Answer (D)

Solution:

Number of ways of selecting 3 green marbles out of 3 green marbles =$^3C_3 = 1$

Number of ways of selecting 3 red marbles out of 4 red marbles = $^4C_3 = 4$

Total ways = 1+4 = 5

Probability =$\frac{5}{^{12}C_3} = 1/44$

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