1000+ Permutation and Combination Questions for Bank PO Pdf - 1

Ans: B

At the first place he can try any of the 4 digits hence in first trial he tries 4 digits.

In the second place he will try 3 remaining digits.

Similarly, he will try 2 and 1 digit at the third and fourth places.

Thus, the number of trials is = 4 × 3 × 2 × 1 = 24.

Ans: D

First question can be answered in 3 ways.

Second questions can be answered in 3 ways.

3 question can be answered in 4 ways.

4 question can be answered in 4 ways.

5 question can be answered in 5 ways.

6 question can be answered in 5 ways.

Hence, by fundamental principle of counting, the required number of sequences of answers

= 3 x 3 x 4 x 4 x 5 x 5 = 3600.

Ans: C

Unit’s place can be filled in 6 ways by any 1 of the digits 1, 2, 3, 4, 5 or 9. Also, ten’s place can be filled in 6 ways.

But hundred’s place can be filled only in 5 ways using either 1, 2, 3, 4, or 5 : 9 cannot be filled in hundred’s place as the required number is < 600.

∴ Required number of numbers = 6 × 6 × 5 = 180.

Ans: D

The word EQUATION has exactly 8 letters which are all different.

∴ Number of words that can be formed = number of permutations of 8 letters taken all at a time

= P(8, 8) = 8!

= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320.

Ans: B

Each apple can be given to any 1 of the 3 boys and this can be done in 3 ways.

∴ The required number of ways = 3⁶ = 729.