The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

Solution

Maths sir · 3 steps

Hand out the guaranteed 2 apples each first, then share the rest with no conditions at all.

1. Meet the minimum. Give every child 2 apples straight away. That uses 3 × 2 = 6 apples and leaves 21 - 6 = 15 identical apples.
2. Free distribution. The remaining 15 apples can go to the three children in any way, with zero allowed. This is the number of non-negative solutions of y₁ + y₂ + y₃ = 15, which stars and bars counts.

   C(15 + 3 - 1, 3 - 1) = C(17, 2) = (17 × 16)/2 = 136

3. So. There are 136 ways, which is option (b).

The idea to keep. For identical objects with a minimum per person, give out the minimums first. Then n objects among r people with zero allowed is C(n+r-1, r-1).

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