The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is :

Solution

Maths sir · 3 steps

Start with seven 1s, which sum to 7, then share the 3 extra units among the digits.

1. Base of all ones. Every digit is at least 1, so seven 1s give a sum of 7. We need 3 more, and each digit can take at most 2 extra, because no digit can go above 3.
2. Split the extra 3. The 3 extra units can be split only as 2 + 1 or as 1 + 1 + 1. A single digit cannot take all 3, since that would make it 4.

   Split of extra 3	Digits in the number	Arrangements
   2 + 1	one 3, one 2, five 1s	7!/5! = 7 × 6 = 42
   1 + 1 + 1	three 2s, four 1s	C(7, 3) = 35

3. So. The total is 42 + 35 = 77. None of the digits is 0, so every arrangement is a genuine seven-digit integer, and the answer is (d).

The idea to keep. When digits have a floor and a ceiling, subtract the floor first. Then share out the leftover sum and list every way to split it, keeping each share under the cap.

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