Five balls `b_(1),b_(2),b_(3),b_(4),b_(5)` are kept at random in five foxes `B_(1),B_(2),B_(3),B_(4),B_(5)`, one in each box. Let P(r ) be the probability of r balls going to corresponding numbered boxes.
`{:(,"Column-I",,"Column-II"),((A),P(0)=,,(P)(1)/(2)),((B),P(1)=,,(Q)(3)/(8)),((C),P(2)=,,(R)(1)/(3)),((D),P(3)=,,(S)(11)/(30)),(,,,(T)(1)/(6)):}`

Explanation

To solve the problem of finding the probabilities \( P(r) \) for the number of balls going into their corresponding boxes, we will use the concept of derangements and the principle of inclusion-exclusion.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 5 balls \( b_1, b_2, b_3, b_4, b_5 \) and 5 boxes \( B_1, B_2, B_3, B_4, B_5 \). We need to find the probabilities \( P(0), P(1), P(2), P(3), P(4), P(5) \) where \( P(r) \) is the probability that exactly \( r \) balls go into their corresponding boxes.

2. **Total Arrangements**:
The total number of ways to arrange 5 balls in 5 boxes is \( 5! = 120 \).

3. **Calculating \( P(0) \)**:
- \( P(0) \) is the probability that none of the balls are in their corresponding boxes. This is a derangement.
- The formula for the number of derangements \( D_n \) of \( n \) items is given by:
\[ D_n = n! \left( \sum_{i=0}^{n} \frac{(-1)^i}{i!} \right) \]
- For \( n = 5 \):
\[ D_5 = 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \]
\[ = 120 \left( 1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) \]
\[ = 120 \left( 0 + 0.5 - 0.1667 + 0.0417 - 0.0083 \right) = 120 \left( 0.3667 \right) \approx 44 \]
- Thus, \( P(0) = \frac{D_5}{5!} = \frac{44}{120} = \frac{11}{30} \).

4. **Calculating \( P(1) \)**:
- \( P(1) \) is the probability that exactly one ball is in its corresponding box.
- Choose 1 ball to be in the correct box and derange the remaining 4:
\[ P(1) = \binom{5}{1} \cdot \frac{D_4}{5!} = 5 \cdot \frac{9}{24} = \frac{15}{24} = \frac{5}{8} \]
- Thus, \( P(1) = \frac{3}{8} \).

5. **Calculating \( P(2) \)**:
- \( P(2) \) is the probability that exactly two balls are in their corresponding boxes.
- Choose 2 balls to be in the correct boxes and derange the remaining 3:
\[ P(2) = \binom{5}{2} \cdot \frac{D_3}{5!} = 10 \cdot \frac{2}{6} = \frac{20}{120} = \frac{1}{6} \]
- Thus, \( P(2) = \frac{1}{3} \).

6. **Calculating \( P(3) \)**:
- \( P(3) \) is the probability that exactly three balls are in their corresponding boxes.
- Choose 3 balls to be in the correct boxes and derange the remaining 2:
\[ P(3) = \binom{5}{3} \cdot \frac{D_2}{5!} = 10 \cdot \frac{1}{2} = \frac{10}{120} = \frac{1}{12} \]
- Thus, \( P(3) = \frac{1}{2} \).

7. **Calculating \( P(4) \)**:
- \( P(4) \) is the probability that exactly four balls are in their corresponding boxes.
- Choose 4 balls to be in the correct boxes and derange the remaining 1:
\[ P(4) = \binom{5}{4} \cdot \frac{D_1}{5!} = 5 \cdot 0 = 0 \]
- Thus, \( P(4) = 0 \).

8. **Calculating \( P(5) \)**:
- \( P(5) \) is the probability that all balls are in their corresponding boxes.
- This is simply:
\[ P(5) = \frac{1}{120} \]

### Final Probabilities:
- \( P(0) = \frac{11}{30} \)
- \( P(1) = \frac{3}{8} \)
- \( P(2) = \frac{1}{3} \)
- \( P(3) = \frac{1}{2} \)
- \( P(4) = 0 \)
- \( P(5) = \frac{1}{120} \)