The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 , 1,2,3,4,5,6,7, such that the sum of their first and last digits should not be more than 8 , is

Solution

Maths sir · 3 steps

Fix the first and last digits together, fill the middle three freely, then throw out 50000 itself.

1. First digit. The number must exceed 50000, so the first digit is 5, 6 or 7. Digits may repeat, because the answer counts numbers like 50000.

   First digit	Last digit allowed (sum ≤ 8)	Choices
   5	0, 1, 2, 3	4
   6	0, 1, 2	3
   7	0, 1	2

2. Middle digits. That gives 4 + 3 + 2 = 9 first-and-last pairs. Each of the three middle places can take any of the 8 digits, so there are 8³ = 512 ways to fill them.

   9 × 512 = 4608

3. So. This count includes 50000, which is not greater than 50000. Removing it leaves 4608 - 1 = 4607, option (d).

The idea to keep. When a condition links two positions, count those positions together as pairs first. Then check the boundary: "greater than" means the boundary number itself is out.

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