Let P be the set of seven digit numbers with sum of their digits equal to 11. If the numbers in P are formed by using the digits 1, 2 and 3 only, then the number of elements in the set P is :

Solution

Maths sir · 3 steps

Start every digit at 1, then share out the 4 extra units as 2s and 3s.

1. Count the extra. Seven digits that are all at least 1 already add up to 7. The remaining 11 - 7 = 4 must come from turning some 1s into 2s (extra 1 each) or 3s (extra 2 each).
2. List the cases. If there are a twos and b threes, then a + 2b = 4. So (a, b) is (4, 0), (2, 1) or (0, 2), and the rest of the seven digits are 1s.

   Digits used	Arrangements	Count
   four 2s, three 1s	7!/(4! 3!)	35
   one 3, two 2s, four 1s	7!/(1! 2! 4!)	105
   two 3s, five 1s	7!/(2! 5!)	21

3. So. None of the digits is 0, so every arrangement is a genuine seven digit number. The total is 35 + 105 + 21 = 161, which is option (c).

The idea to keep. When digits are restricted to a small set, fix the counts of each digit from the sum condition first, then arrange each case with n!/(p! q! r!).

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