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# Permutation and Combination Questions for SBI Clerk Set-2 PDF

Download SBI Clerk Permutation and Combination Questions & Answers Set-2 PDF for SBI Clerk Prelims and Mains exam. Very Important SBI Clerk Permutation and Combination Questions with solutions.

Question 1: 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

Question 2: 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

Question 3: 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

Question 4: 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

Question 5: 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

Question 6: 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

Question 7: Let x=123456….424344. What is the remainder when x is divided by 45?

a) 0

b) 9

c) 4

d) none

Instructions

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

Question 8: 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

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 9: 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

Question 10: The committee should have 2 trainees and 3 research associates ?

a) 15

b) 45

c) 60

d) 9

e) None of these

Question 11: A die is thrown twice. What is the probability of getting a sum 7 from both the throws?

a) $\frac{5}{18}$

b) $\frac{1}{18}$

c) $\frac{1}{9}$

d) $\frac{1}{6}$

e) $\frac{5}{36}$

Question 12: 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

Question 13: In how many different ways can the letters of the word BAKERY be arranged ?

a) 2400

b) 2005

c) 720

d) 5040

e) None of these

Question 14: In how many different ways can the letters of the word DAILY be arranged ?

a) 60

b) 48

c) 160

d) 120

e) None of these

Question 15: In how many different ways can the letters of the word ‘PARTY’ be arranged ?

a) 120

b) 2005

c) 2400

d) 720

e) None of these

Answers & Solutions:

1) Answer (D)

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

2) Answer (A)

Circular permutation = n! (n – 1)!

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

3) Answer (B)

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

4) Answer (B)

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.

5) Answer (E)

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.

6) Answer (A)

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.

7) Answer (B)

To check if a number is divisible by 45, we have to check the divisibility of the number by 5 and 9.
x = 123456789………424344
Let us check the divisibility of x by 5:
For a number to be divisible by 5, it’s unit digit has to be 0 or 5.
Since the unit digit is 4, the remainder obtained when x is divided by 5 is 4
Let us check the divisibility of x by 9:

Sum from 1 to 9 = 45. Sum from 10 to 19 = 45+10 =55. Sum from 20 to 29 =65. Sum from 30 to 39 = 75. Sum from 40 to 44 = 30. Therefore total sum of the digits of the number = 270. Therefore, the number is divisible by 9.

Using Chinese Remainder theorem,
5p+4=9q
When p = 1, both LHS = RHS
Hence when p = 1, 5p+4=9
So 9 is the remainder when x is divided by 45.

8) Answer (D)

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

9) Answer (A)

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

10) Answer (C)

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

11) Answer (D)

The number 7 can be obtained in 6 ways. (6,1), (1,6), (5,2), (2,5), (4,3), (3,4).

Required probability = 6/36 = 1/6

12) Answer (D)

There are 6 letters in total of which three are repeated.

Hence, total number of arrangements = $\frac{6!}{3!}$ = 120

13) Answer (C)

As there are 6 different letters in the word “Bakery” hence, total number of ways of arranging it will be = 6! = 720

14) Answer (D)

If a word has n different letters then they can be arranged in n! different ways.
DAILY has 5 different letters which can bve arranged in 5! different ways.
5!=5*4*3*2*1=120
Therefore, option D is answer.

15) Answer (A)

The word “Party” can be arranged in 5! ways as there are all different letters in the word.

Hence, answer will be = 120

We hope this Permutation and Combination Questions & Answers PDF of SBI Clerk is very Useful for preparation of SBI Clerk Exams.