In how many ways can we distribute 9 pencils to 4 children so that each child…

Concept: For distributing r identical objects among n distinct groups so that every group receives at least one object, the number of ways equals C(r-1, n-1) — obtained by lining up the r objects in a row and choosing (n-1) of the (r-1) internal gaps to place dividers, since the two end gaps cannot be used without leaving some group empty. This is the classic stars-and-bars theorem for distributions with a minimum-one-per-group constraint.

Application:

1. Here, r = 9 pencils and n = 4 children, each of whom must receive at least one pencil.

2. By the stars-and-bars formula for at-least-one distributions, the number of ways = C(r-1, n-1) = C(9-1, 4-1) = C(8, 3).

3. Expand C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.

Cross-check: Independent check: first give each of the 4 children 1 pencil (using 4 of the 9 pencils), leaving 5 pencils to be distributed freely (0 or more each) among the 4 children. The number of ways to do this is C(5 + 4 - 1, 4 - 1) = C(8, 3) = 56 — the same result, confirming the answer.

Result: So there are 56 ways to distribute the 9 pencils among the 4 children.