In the previous question (no.63) find the number of ways in which only two balls can be put in the correct boxes i.e., the colour of box and the colour of contained ball be same.

Options

A. A) 3
B. B) 6
C. C) 9
D. D) none of these

Answer

To solve the problem of finding the number of ways in which exactly two balls can be put in the correct boxes, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 4 different colored balls (Red, Green, Blue, White) and 4 boxes of the same colors. We need to find the number of ways to place exactly 2 balls in their correct boxes while the other 2 balls must be placed in incorrect boxes.

2. **Choosing the Correct Balls**: We first need to select 2 balls that will be placed in their correct boxes. The number of ways to choose 2 balls from 4 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Here, \( n = 4 \) (total balls) and \( r = 2 \) (balls to be placed correctly).
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

3. **Arranging the Incorrect Balls**: After placing 2 balls correctly, we have 2 balls left that need to be placed in the incorrect boxes. For the remaining 2 balls, they can only be placed in each other's boxes. There is only 1 way to do this (i.e., if Ball A goes into Box B, then Ball B must go into Box A).

4. **Calculating Total Arrangements**: Since we have 6 ways to choose which 2 balls are placed correctly and 1 way to arrange the remaining 2 balls incorrectly, the total number of arrangements is:
\[ \text{Total Ways} = \text{Ways to choose 2 correct balls} \times \text{Ways to arrange incorrect balls} = 6 \times 1 = 6 \]

5. **Final Answer**: Therefore, the total number of ways in which exactly two balls can be put in the correct boxes is **6**.