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

<p>To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls and bins"), which is a way to solve problems involving distributing identical items into distinct groups with certain restrictions.</p> <p>First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That accounts for 6 apples (2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to distribute freely among the three children.</p> <p>The "stars and bars" technique involves representing the apples as stars (*) and the divisions between children as bars (|). For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as **|*|**. This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.</p> <p>In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.</p> <p>The total number of objects we're arranging is 15 apples + 2 bars = 17 objects. We need to choose 2 positions out of these 17 to place the bars. The remaining positions will be occupied by the stars (apples).</p> <p>The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:</p> $$ \text{Number of ways} = \binom{17}{2} = \frac{17!}{2!(17-2)!} = \frac{17 \times 16}{2 \times 1} = 136 $$ <p>Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child gets at least 2 apples. The correct answer is Option B: 136.</p>