A bag contains n white and n black balls. Pairs of balls are drawn without replacement until the bag is empty. If the number of ways in which each pair consists of one black and one white ball is 576, then n is

Options

A. 4
B. less than 5
C. greater than 24
D. less than 12

Answer

To solve the problem, we need to determine the value of \( n \) given that the number of ways to draw pairs of balls (one white and one black) from a bag containing \( n \) white and \( n \) black balls is 576.

### Step-by-step Solution:

1. **Understanding the Pair Drawing Process**:
- We have \( n \) white balls and \( n \) black balls.
- We draw pairs of one white and one black ball until the bag is empty.
- This means we will perform \( n \) draws in total.

2. **Calculating the Number of Ways to Draw Pairs**:
- For the first pair, we can choose 1 white ball from \( n \) and 1 black ball from \( n \):
\[ \text{Ways for first pair} = n \times n = n^2 \]
- For the second pair, we have \( n-1 \) white balls and \( n-1 \) black balls left:
\[ \text{Ways for second pair} = (n-1) \times (n-1) = (n-1)^2 \]
- Continuing this process, the number of ways to draw the pairs will be:
\[ n^2 \times (n-1)^2 \times (n-2)^2 \times \ldots \times 1^2 \]

3. **Expressing the Total Number of Ways**:
- The total number of ways to draw all pairs can be expressed as:
\[ n^2 \times (n-1)^2 \times (n-2)^2 \times \ldots \times 1^2 = (n!)^2 \]
- This is because \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \).

4. **Setting Up the Equation**:
- We know that the total number of ways is equal to 576:
\[ (n!)^2 = 576 \]

5. **Taking the Square Root**:
- Taking the square root of both sides gives:
\[ n! = \sqrt{576} = 24 \]

6. **Finding \( n \)**:
- Now we need to find \( n \) such that \( n! = 24 \).
- The factorial values are:
- \( 1! = 1 \)
- \( 2! = 2 \)
- \( 3! = 6 \)
- \( 4! = 24 \)
- Therefore, \( n = 4 \).

### Conclusion:
The value of \( n \) is \( 4 \).