The number of ways of distributing 5 identical balls in into three boxes so that no box is empty (each box being large enough to accommodate all balls), is

Options

A. `3^(5)`
B. `5^(3)`
C. 15
D. 6

Explanation

To solve the problem of distributing 5 identical balls into 3 boxes such that no box is empty, we can follow these steps:

### Step 1: Understand the problem
We need to distribute 5 identical balls into 3 boxes (let's call them Box 1, Box 2, and Box 3) with the condition that each box must contain at least one ball.

### Step 2: Adjust for the condition of non-empty boxes
Since each box must contain at least one ball, we can start by placing one ball in each box. This ensures that no box is empty.

After placing one ball in each box, we have:
- 3 balls (1 in each box) already placed
- 2 balls remaining to be distributed

### Step 3: Formulate the equation
Now, we need to distribute the remaining 2 balls into the 3 boxes. The new equation we need to solve is:
\[ x_1 + x_2 + x_3 = 2 \]
where \( x_1, x_2, \) and \( x_3 \) are the number of additional balls in Box 1, Box 2, and Box 3 respectively.

### Step 4: Use the stars and bars method
The problem of distributing \( n \) identical items (balls) into \( r \) distinct groups (boxes) can be solved using the "stars and bars" theorem. The formula for the number of ways to distribute \( n \) identical items into \( r \) distinct groups is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 2 \) (remaining balls) and \( r = 3 \) (boxes).

### Step 5: Apply the formula
Plugging in the values, we have:
\[ \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} \]

### Step 6: Calculate the binomial coefficient
Now, we calculate \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

### Conclusion
Thus, the number of ways to distribute 5 identical balls into 3 boxes such that no box is empty is **6**.