There are 5 different boxes and 7 different balls. All the balls are to be distributed in the 5 boxes placed in a row so that any box can receive any number of balls.
Suppose all the balls are identical, then in how many ways can all these balls be distributed into these boxes?

Options

A. A) 110
B. B) 220
C. C) 330
D. D) 440

Answer

To solve the problem of distributing 7 identical balls into 5 different boxes, we can use the "stars and bars" theorem in combinatorics. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have 7 identical balls and 5 different boxes. We need to find out how many ways we can distribute these balls into the boxes, where each box can hold any number of balls (including zero).

### Step 2: Apply the Stars and Bars Theorem
The stars and bars theorem states that if we want to distribute \( n \) identical items (stars) into \( r \) distinct groups (boxes), the number of ways to do this is given by the formula:

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

Where:
- \( n \) = number of identical items (balls)
- \( r \) = number of distinct groups (boxes)

### Step 3: Substitute the Values
In our case:
- \( n = 7 \) (the number of balls)
- \( r = 5 \) (the number of boxes)

Using the formula, we substitute these values:

\[ \text{Number of ways} = \binom{7 + 5 - 1}{5 - 1} = \binom{11}{4} \]

### Step 4: Calculate \( \binom{11}{4} \)
Now we need to calculate \( \binom{11}{4} \):

\[ \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4! \cdot 7!} \]

This simplifies to:

\[ \binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} \]

### Step 5: Perform the Calculation
Calculating the numerator:

\[ 11 \times 10 = 110 \]
\[ 110 \times 9 = 990 \]
\[ 990 \times 8 = 7920 \]

Now calculating the denominator:

\[ 4 \times 3 \times 2 \times 1 = 24 \]

Now divide the numerator by the denominator:

\[ \frac{7920}{24} = 330 \]

### Final Answer
Thus, the total number of ways to distribute 7 identical balls into 5 different boxes is **330**.

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