The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Options

A. `""^(8)C_(3)`
B. 21
C. `3^8`
D. 5

Answer

To solve the problem of distributing 8 identical balls into 3 distinct boxes such that none of the boxes is empty, we can follow these steps:

### Step 1: Understand the Problem
We need to distribute 8 identical balls into 3 distinct boxes, ensuring that each box contains at least one ball.

### Step 2: Use the Stars and Bars Theorem
To ensure that no box is empty, we can first place one ball in each box. This guarantees that all boxes will have at least one ball.

### Step 3: Distribute Remaining Balls
After placing one ball in each of the 3 boxes, we have used 3 balls. Therefore, we have:
\[ 8 - 3 = 5 \]
balls left to distribute freely among the 3 boxes.

### Step 4: Apply Stars and Bars
Now, we need to find the number of ways to distribute these 5 identical balls into 3 distinct boxes. According to the stars and bars theorem, 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:
- \( n = 5 \) (the remaining balls)
- \( r = 3 \) (the boxes)

Thus, we need to calculate:
\[ \binom{5 + 3 - 1}{3 - 1} = \binom{7}{2} \]

### Step 5: Calculate the Binomial Coefficient
Now, we calculate \( \binom{7}{2} \):
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21 \]

### Final Answer
The number of ways to distribute 8 identical balls into 3 distinct boxes such that none of the boxes is empty is **21**.

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