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

The Correct Option is D

To find the number of ways to distribute 21 identical apples among three children such that each child gets at least 2 apples, we can solve the problem using the "stars and bars" theorem. This is a classic example of a combinatorics problem where we need to distribute indistinguishable objects into distinguishable bins with certain restrictions.

First, assign 2 apples to each child to meet the condition that each child gets at least 2 apples. Therefore, we distribute:

• Apples given initially to each child: \(3 \times 2 = 6\) apples

Now, we have \(21 - 6 = 15\) apples left to distribute among the 3 children with no further restrictions.

According to the stars and bars method, the problem now is equivalent to finding the number of non-negative integer solutions to the equation:

\(x_1 + x_2 + x_3 = 15\)

where \(x_1\), \(x_2\), and \(x_3\) are the number of additional apples given to the first, second, and third child, respectively.

The number of solutions is given by the formula for combinations with repetition, which is:

\(\binom{n+k-1}{k-1}\)

In our case, \(n = 15\) (apples left) and \(k = 3\) (children), so:

\(\binom{15+3-1}{3-1} = \binom{17}{2}\)

Calculate \(\binom{17}{2}\) as follows:

\(\binom{17}{2} = \frac{17 \times 16}{2 \times 1} = \frac{272}{2} = 136\)

Thus, the number of ways to distribute the apples under the given conditions is 136.

Therefore, the correct answer is 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 $136$.