Question

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

• A. 406
• B. 130
• C. 142
• D. 136

Answer 1

Answer: (D)

136

Answer 2

To ensure each child gets at least 2 apples, we can start by giving 2 apples to each of the three children.

Total apples given:
$2 \times 3 = 6$

Remaining apples:
$21 - 6 = 15$

Now, we need to distribute these remaining 15 apples among the 3 children with no additional restrictions (each child can get zero or more apples).

This problem now becomes a "distribution of identical items into distinct groups" problem.

We can use the stars and bars method to calculate the number of ways to distribute 15 identical apples among 3 children.

The formula for distributing $n$ identical items into $r$ distinct groups is:
$n + (r-1)_{C_{r-1}}$

Here, $n = 15$ (remaining apples) and $r = 3$ (children), so:

$15 + (3-1)_{C_{3-1}} = ^{17}C_{2}$

Now, calculating $^{17}C_{2}$:

$^{17}C_{2} = \frac{17 \times 16}{2} = 136$

Thus, the number of ways to distribute the 21 apples such that each child receives at least 2 apples is $136$.