In how many ways can 10 balls be divided between two boys, one receiving two and the other eight balls

Answer

To solve the problem of how many ways 10 balls can be divided between two boys, one receiving 2 balls and the other receiving 8 balls, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Total Number of Balls and Distribution**:
We have a total of 10 balls. One boy (let's say Boy A) will receive 2 balls, and the other boy (Boy B) will receive 8 balls.

2. **Case 1: Boy A receives 2 balls**:
We need to calculate how many ways Boy A can choose 2 balls from the 10 available. This can be calculated using the combination formula \( \binom{n}{r} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose.
\[ \text{Ways for Boy A} = \binom{10}{2} \]

3. **Calculate \( \binom{10}{2} \)**:
Using the combination formula:
\[ \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]

4. **Case 2: Boy B receives 2 balls**:
Alternatively, we can consider the case where Boy B receives 2 balls and Boy A receives 8 balls. The number of ways Boy B can choose 2 balls from the 10 available is:
\[ \text{Ways for Boy B} = \binom{10}{2} \]
Since this is the same calculation, we again find:
\[ \binom{10}{2} = 45 \]

5. **Combine the Two Cases**:
Since both cases are mutually exclusive (Boy A cannot receive 2 balls while Boy B receives 2 balls at the same time), we add the number of ways from both cases:
\[ \text{Total Ways} = \binom{10}{2} + \binom{10}{2} = 45 + 45 = 90 \]

### Final Answer:
Thus, the total number of ways to divide the 10 balls between the two boys is **90**.

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