Animated Solution for Mathematics - Permutations and Combinations: 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

Distributing Identical Items

Welcome, future engineer! Today, we are diving into a classic problem that sits at the heart of combinatorics. It might seem like a simple task of handing out apples, but beneath the surface lies a beautiful mathematical structure.

We have 21 identical apples and 3 children, and we need to distribute them such that each child gets at least 2 apples. Let us embark on this journey together.

Visualizing the Problem

Imagine you are standing in front of three children, holding a basket of 21 identical apples. You want to distribute them, but there is a rule: no child can be left with fewer than 2 apples.

Let $x_1,x_2,x_3$ be the number of apples given to the first, second, and third child, respectively. Since we are distributing all 21 apples, we have the equation:

The constraint is $x_i\ge 2$ for each child. This is the core of our problem.

The Pre-distribution Strategy

The secret to handling this inequality is to simplify the problem by satisfying the constraint first. If each child must have at least 2 apples, let us just give them those apples right now.

We have 3 children, and each needs 2 apples. That is $3\times 2=6$ apples gone from our basket.

Now, how many apples are left? We started with 21, and we gave away 6, so $21-6=15$ apples remain. By handling the minimum requirement upfront, we have transformed a constrained problem into an unconstrained one.

The Transformation

Let us formalize this. We introduce new variables, $y_1,y_2,y_3$, which represent the extra apples each child receives beyond the 2 they already have. So, $x_i=y_i+2$, where $y_i\ge 0$.

Substituting this into our original equation, we get:

Simplifying this, we get:

Now, we are looking for the number of non-negative integer solutions to this equation. This is a much friendlier problem!

The Stars and Bars Method

This is where the Stars and Bars method shines. This theorem tells us that the number of ways to distribute $n$ identical items into $r$ distinct groups is given by the formula:

In our case, $n=15$ (the remaining apples) and $r=3$ (the children). Plugging these values into our formula, we get:

The Final Calculation

Now, we just need to compute $\left(\genfrac{}{}{0ex}{}{17}{2}\right)$. This is calculated as:

We can simplify this by dividing 16 by 2 to get 8, so we are left with $17\times 8$. Calculating this, we get 136.

There it is! The number of ways to distribute the apples is 136. Remember, whenever you face a problem with identical items and minimum constraints, just pre-allocate the minimums and let the Stars and Bars formula do the heavy lifting. You have got this!