The number of ways to distribute 30 identical candies among four children C₁, C₂, C₃ and C₄ so that C₂ receives at least 4 and at most 7 candies, C₃ receives at least 2 and at most 6 candies, is equal to :

Solution

Maths sir · 3 steps

Fix the two restricted children first; then C₁ and C₄ share whatever is left, with no limits.

1. Fix the restricted ones. Let C₂ get a candies with 4 ≤ a ≤ 7, and C₃ get b with 2 ≤ b ≤ 6. That gives only 4 × 5 = 20 pairs (a, b) to handle.
2. Share the rest. The remaining 30 - a - b identical candies go to C₁ and C₄ with no restriction. Splitting n identical items between two people can be done in n + 1 ways, so this gives 31 - a - b ways.

   N = Σₐ₌₄⁷Σ_(b=2)⁶ (31 - a - b)

   a (for C₂)	Sum over b = 2 to 6	Ways
   4	25 + 24 + 23 + 22 + 21	115
   5	24 + 23 + 22 + 21 + 20	110
   6	23 + 22 + 21 + 20 + 19	105
   7	22 + 21 + 20 + 19 + 18	100

3. So. N = 115 + 110 + 105 + 100 = 430, which is option (d).

The idea to keep. When only a few children have both a lower and an upper limit, loop over their small ranges. The free children then share the rest in n + 1 ways (two people) or C(n+r-1, r-1) ways (r people).

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