Find the number of ways in which five identical balls can be distributed among ten identical boxes, if not more than one can go into a box.

Answer

To solve the problem of distributing five identical balls into ten identical boxes with the condition that no box can contain more than one ball, we can follow these steps:

### Step 1: Understand the Problem
We have 5 identical balls and 10 identical boxes. Since no box can contain more than one ball, we can only place a maximum of 5 balls in 5 different boxes.

### Step 2: Identify the Combinations
Since the boxes are identical, the arrangement of the boxes does not matter. We only need to select which 5 boxes out of the 10 will contain the balls.

### Step 3: Calculate the Number of Ways
The number of ways to choose 5 boxes from 10 can be calculated using the combination formula:
\[ \text{Number of ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of boxes (10) and \( r \) is the number of boxes to choose (5).

### Step 4: Substitute the Values
Substituting the values into the combination formula:
\[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5! \cdot 5!} \]

### Step 5: Simplify the Factorials
Now, we can simplify:
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \]
Thus,
\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5!} \]
Calculating \( 5! \):
\[ 5! = 120 \]
Now substituting back:
\[ \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120} \]

### Step 6: Calculate the Numerator
Calculating the numerator:
\[ 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
\[ 5040 \times 6 = 30240 \]

### Step 7: Final Calculation
Now divide by \( 120 \):
\[ \frac{30240}{120} = 252 \]

### Conclusion
Thus, the number of ways to distribute five identical balls into ten identical boxes, with no box containing more than one ball, is **252**.

---