The number of five digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits 0,1,3,5,7 and 9 without repetition, is equal to :

Solution

Maths sir · 4 steps

Fix the two restricted places first, the first digit and the last, then fill the middle three.

1. Restricted places. To be greater than 40000, the first digit must be 5, 7 or 9. To be divisible by 5, the last digit must be 0 or 5. The digit 5 appears in both lists, so split into two cases by the last digit.
2. Last digit 0. The first digit can be 5, 7 or 9, which gives 3 choices. That leaves 4 digits for the three middle places, so 4 × 3 × 2 = 24 ways. Total: 3 × 24 = 72.
3. Last digit 5. The 5 is used up, so the first digit can only be 7 or 9, which gives 2 choices. Again 4 digits remain for the middle, giving 24 ways. Total: 2 × 24 = 48.
4. So. The total is 72 + 48 = 120, which is option (c).

The idea to keep. When one digit can sit in two restricted places, split into cases on that digit, so you never use it twice. Always fill the restricted places before the free ones.

Answer from the NTA final answer key. Solution by PrepWiser.