There are two bosy `B_(1)` and `B_(2)`. `B_(1)` and `n_(1)` different toys and `B_(2)` and `n_(2)` different toys. Find the number of ways in which `B_(1)` and `B_(2)` can exchange their toys in such a way that after exchanging they still have same number of toys but not the same set.

Answer

To solve the problem of how many ways two boys \( B_1 \) and \( B_2 \) can exchange their toys while ensuring they still have the same number of toys but not the same set, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( n_1 \) be the number of different toys that boy \( B_1 \) has.
- Let \( n_2 \) be the number of different toys that boy \( B_2 \) has.

2. **Determine the Exchange Possibilities**:
- The boys can exchange toys in various ways. They can exchange one toy, two toys, etc., up to the minimum of \( n_1 \) and \( n_2 \).

3. **Calculate the Number of Ways to Exchange Toys**:
- If they exchange \( k \) toys, boy \( B_1 \) will give \( k \) toys to boy \( B_2 \) and receive \( k \) toys from boy \( B_2 \).
- The number of ways to choose \( k \) toys from \( n_1 \) is given by \( \binom{n_1}{k} \).
- The number of ways to choose \( k \) toys from \( n_2 \) is given by \( \binom{n_2}{k} \).
- Therefore, the total number of ways to exchange \( k \) toys is \( \binom{n_1}{k} \times \binom{n_2}{k} \).

4. **Sum Over All Possible Exchanges**:
- Since they can exchange from \( k = 1 \) to \( k = \min(n_1, n_2) \), we need to sum the number of ways for each possible \( k \):
\[ \text{Total Ways} = \sum_{k=1}^{\min(n_1, n_2)} \binom{n_1}{k} \times \binom{n_2}{k} \]

5. **Adjust for Same Set Condition**:
- The problem specifies that after the exchange, they should not have the same set of toys. This means we need to subtract the cases where they end up with the same set of toys.
- The only way they can end up with the same set of toys is if they exchange all toys. Therefore, we need to subtract the case when \( k = \min(n_1, n_2) \):
\[ \text{Final Count} = \sum_{k=1}^{\min(n_1, n_2) - 1} \binom{n_1}{k} \times \binom{n_2}{k} \]

### Final Answer:
Thus, the number of ways in which \( B_1 \) and \( B_2 \) can exchange their toys while ensuring they still have the same number of toys but not the same set is given by:
\[ \sum_{k=1}^{\min(n_1, n_2) - 1} \binom{n_1}{k} \times \binom{n_2}{k} \]