IIM B UG (B.Sc-Hons.)- Sample Paper 2026 · Quant & DI · Question 19

The even digits available are 0, 2, 4, 6 and the odd digits are 1, 3, 5.

Exactly two of the five digits are even, so three are odd — and since only three odd digits exist, all of 1, 3 and 5 must be used.

To be divisible by 5 the number must end in 0 or 5. Take the two cases separately.

Case 1: the number ends in 0.

Then 0 is one of the two even digits, and the other is chosen from 2, 4, 6 in 3 ways.

The first four places hold 1, 3, 5 and that even digit, in 4! = 24 orders, and no leading-zero problem arises because 0 is at the end.

Case 1 gives 3 × 24 = 72 numbers.

Case 2: the number ends in 5.

The two even digits are chosen from 0, 2, 4, 6, and the first four places hold 1, 3 and those two evens.

If 0 is one of them (3 ways to pick the other), the 4 digits arrange in 4! = 24 ways, of which 3! = 6 start with 0, leaving 18. That gives 3 × 18 = 54.

If 0 is not chosen, the two evens come from 2, 4, 6 in 3 ways and all 24 arrangements are valid, giving 3 × 24 = 72.

Case 2 gives 54 + 72 = 126 numbers.

Total = 72 + 126 = 198.