The number of ways, in which 16 oranges can be distributed to four children such that each child get

Q: Wait, why can't we just use permutations? Aren't the children distinct?

That is a fantastic question! You are right that the children are distinct, but the oranges are identical. If the oranges were distinct, we would use $4^{16}$. Since the oranges are identical, we only care about the *count* each child receives, not *which* orange they get. That is why we use the Stars and Bars method, which is specifically designed for identical items.

Q: Why do we subtract 1 from each child? Does it change the total?

Great observation! We are not changing the total number of ways; we are just changing the *perspective*. By giving each child one orange first, we satisfy the 'at least one' condition immediately. This transforms the problem from $x_i \ge 1$ to $y_i \ge 0$, which is the standard form for the Stars and Bars formula. It is a mathematical shortcut, not a change to the problem's core.

Q: Is there a formula for this? I keep forgetting if it's $\binom{n+r-1}{r-1}$ or $\binom{n-1}{r-1}$.

It is easy to get confused! Remember the logic: you have $n$ stars and $r-1$ bars. The total number of positions is $n + (r-1)$. You need to choose the positions for the bars (or the stars). So it is $\binom{n+r-1}{r-1}$. If you have the 'at least one' constraint, you can also use $\binom{n-1}{r-1}$ directly. Both lead to the same result!

Question

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

Options:

(A) 455

(B) 429

(C) 403

(D) 484

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of 16 identical oranges, and you have 4 children waiting eagerly. Your task is to distribute these oranges such that every single child gets at least one.

The number of possibilities seems overwhelming at first glance. However, combinatorics allows us to build a logical framework to solve this without listing every single case.

### The Constraint as a Gift

The problem states that each child must receive at least one orange. Let $x_1, x_2, x_3, x_4$ be the number of oranges each child receives.

Our equation is:
$$x_1 + x_2 + x_3 + x_4 = 16$$
subject to the condition $x_i \ge 1$.

To simplify this, we pre-allocate one orange to each child. We take 4 oranges out of our pile of 16 and hand one to each child, ensuring the condition is satisfied.

We now have $16 - 4 = 12$ oranges left to distribute. Let $y_i$ be the *additional* oranges each child receives.

Our new equation becomes:
$$y_1 + y_2 + y_3 + y_4 = 12$$
where $y_i \ge 0$. The constraint is now much simpler, as each child can receive zero or more additional oranges.

### The Magic of Stars and Bars

We now enter the realm of the **Stars and Bars** method. We have 12 identical items to distribute into 4 distinct bins.

Imagine the 12 oranges as stars in a line: $\star \star \star \star \star \star \star \star \star \star \star \star$. To divide these into 4 groups, we need 3 separators, or bars.

For example, $\star \star | \star \star \star | \star \star \star \star \star | \star \star$ represents a distribution where the children get 2, 3, 5, and 2 additional oranges respectively. The total number of positions in our line is the number of stars plus the number of bars:
$$12 + 3 = 15$$

We simply need to choose 3 positions out of these 15 to place our bars. The number of ways to do this is given by the combination formula $\binom{15}{3}$.

### The Final Calculation

The formula for combinations is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. Applying this to our values:

$$\binom{15}{3} = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1}$$

Simplifying this expression:

$$\binom{15}{3} = 5 	imes 7 	imes 13 = 455$$

The total number of ways to distribute the oranges is **455**. Through this transformation, we have reduced a complex distribution problem into a classic combinatorial arrangement.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

Select Answer:

Visualized Solution

Visualizing the Problem

Defining the Variables

The 'At Least One' Transformation

Simplified Equation

Applying Stars and Bars

Formula Application

Final Calculation

Direct Formula (Alternative)

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Constraint as a Gift

The Magic of Stars and Bars

The Final Calculation

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