In how many ways can 6 apples be distributed among 4 boys, there being no restriction to the number of apples each boy may get?

Options

A. 6729
B. 5739
C. 7592
D. 4096

Answer

To solve the problem of distributing 6 apples among 4 boys with no restrictions on the number of apples each boy can receive, we can use the "stars and bars" theorem from combinatorics. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We need to distribute 6 identical apples (stars) among 4 distinct boys (bins). Since there are no restrictions, any boy can receive any number of apples, including zero.

**Hint**: Visualize the apples as stars and the separations between different boys as bars.

### Step 2: Apply the Stars and Bars Theorem
According to the stars and bars theorem, if we have \( n \) identical items (apples) to distribute among \( k \) distinct groups (boys), the number of ways to do this is given by the formula:

\[ \text{Number of ways} = \binom{n + k - 1}{k - 1} \]

In our case:
- \( n = 6 \) (the number of apples)
- \( k = 4 \) (the number of boys)

### Step 3: Substitute the Values into the Formula
Now we substitute \( n \) and \( k \) into the formula:

\[ \text{Number of ways} = \binom{6 + 4 - 1}{4 - 1} = \binom{9}{3} \]

### Step 4: Calculate the Binomial Coefficient
Now, we need to calculate \( \binom{9}{3} \):

\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!} \]

Calculating the factorials:

\[ 9! = 9 \times 8 \times 7 \times 6! \]
Thus, we can simplify:

\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Step 5: Conclusion
Therefore, the number of ways to distribute 6 apples among 4 boys is **84**.

**Final Answer**: 84

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