Permutation and Combination Questions for SBI Clerk

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Permutation and Combination Questions for SBI Clerk
Permutation and Combination Questions for SBI Clerk

Permutation and Combination Questions for SBI Clerk:

Download permutation and combinations questions and answers. Useful for banking exams of SBI Po and SBI clerk.

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Question 1: Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?

a) 4320
b) 720
c) 2160
d) 120
e) None of these

Question 2: Certain number of pieces of an article are to be packed in boxes, such that each box contains 145 pieces. If after packing them in 32 boxes 25 pieces are left out, what was the number of pieces to be packed ?

a) 4566
b) 4655
c) 4465
d) 4640
e) None of these

Question 3: How many such pairs of letters are there in the word FOREHAND each of which have as many letters between them in the word as they have in the English alphabet?

a) None
b) One
c) Two
d) Three
e) More than three

Question 4: If it is possible to make only one meaningful word from the second, the fourth, the sixth and the ninth letters of the word PROACTIVE, using each letter only once, second letter of that word is your answer. If more than one word can be formed your answer is M and if no such word can be formed your answer is N.

a) A
b) E
c) T
d) M
e) N

Question 5: In how many different ways can the letters of the word DRASTIC be arranged in such a way that the vowels always come together ?

a) 720
b) 360
c) 1440
d) 540
e) None of these

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Question 6: In how many different ways can the letters of the word ‘FRANCE’ be arranged?

a) 2400
b) 720
c) 2005
d) 5040
e) None of these

Question 7: In how many different ways can the letters of the word ‘CREAM’ be arranged?

a) 720
b) 240
c) 360
d) 504
e) None of these

Question 8: In how many different ways can the letters of the word “ARISE’ be arranged?

a) 90
b) 60
c) 180
d) 120
e) None of these

Question 9: In how many different ways can the letters of the world ‘SECOND’ be arranged?

a) 720
b) 120
c) 5040
d) 270
e) None of these

Question 10: In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red ?

a) 29/35
b) 7/15
c) 23/35
d) 2/5
e) 19/35

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Solutions for Permutation and Combination Questions for SBI Clerk:

Solutions:

1) Answer (A)

Word – DIRECTOR
So “I,E,O” are there are 3! ways to arrange the vowels
Now “D,R,C,T,R” are the remaining alphabets ,
Condition is that the vowels should always be together so we can assume the vowels as a single alphabet/unit say “X” (‘X’=’I,E,O’) so now we have a new word – “D,R,C,T,R,X”
Possible arrangements for this word = 6!
Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of ‘DRCTRX’) $$\times$$ (Possible arrangements of vowels)
= 6! $$\times$$ 3! = $$720 \times 6 = 4320$$
=> Ans – (A)

2) Answer (E)

No. of pieces
= 32 * 145 + 25
= 4640 + 25 = 4665

3) Answer (C)

Word = FOREHAND
There are 2 pairs of letters which have as many letters between them in the word as they have in the English alphabet
FA and RN

4) Answer (D)

Word = PROACTIVE
2nd, 4th, 6th and 9th letters = R, A, T, E
No. of words that can be formed by using (R,A,T,E)
= Rate , Tear
Since only 2 words are formed
=> Ans = M

5) Answer (C)

There are 7 letters in the word ‘DRASTIC’ including 2 vowels (A,I) and 5 consonants (D,R,S,T,C).
Considering the two vowels as 1 letter, we have 6 letters which can be arranged in 6! ways
But  corresponding to each way of these arrangements, the vowels can be put together in 2! ways.
Hence, required number of words = $$6! \times 2!$$
= 720 * 2 = 1440

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6) Answer (B)

The word FRANCE consists of 6 distinct letters
=> Required number of arrangements = 6!
= 720

7) Answer (E)

The word CREAM consists of five distinct letters
=> No. of ways the letters will be arranged = 5!= 120

8) Answer (D)

The word ‘ARISE’ has 5 different letters
=> Required number of arrangements = $$5!$$
= $$5 \times 4 \times 3 \times 2 \times 1 = 120$$

9) Answer (A)

The word ‘SECOND’ has 6 different letters
=> Required number of arrangements = $$6!$$
= $$6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$

10) Answer (C)

Probability that at least 1 ball is red = 1 – probability that none of them is red.
Probability that none if the two balls is red = (9/15)(8/14)
Probability that at least 1 ball is red = 1 – probability that none of them is red. = 1- [(9/15)(8/14)] = (210-72)/210
= 138/210
=23/35
Option C is the correct answer.

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