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 1: Understand the Problem
We have two boys:
- \( B_1 \) has \( n_1 \) different toys.
- \( B_2 \) has \( n_2 \) different toys.

They want to exchange toys such that after the exchange, they still have the same number of toys, but the sets of toys they have are different.

### Step 2: Define the Exchange
Let \( k \) be the number of toys exchanged. After the exchange, both boys will have \( n_1 - k \) toys from their original set and \( k \) toys from the other boy's set.

### Step 3: Choose the Toys to Exchange
For each possible value of \( k \) (where \( k \) can range from 1 to the minimum of \( n_1 \) and \( n_2 \)), we need to choose:
- \( k \) toys from \( B_1 \) (which has \( n_1 \) toys).
- \( k \) toys from \( B_2 \) (which has \( n_2 \) toys).

The number of ways to choose \( k \) toys from \( B_1 \) is given by \( \binom{n_1}{k} \), and the number of ways to choose \( k \) toys from \( B_2 \) is given by \( \binom{n_2}{k} \).

### Step 4: Sum Over All Possible Values of \( k \)
We need to sum the products of these combinations for all possible values of \( k \):
\[ \text{Total Ways} = \sum_{k=1}^{\min(n_1, n_2)} \binom{n_1}{k} \cdot \binom{n_2}{k} \]

### Step 5: Use the Hockey Stick Identity
Using the hockey stick identity in combinatorics, we can simplify the summation:
\[ \sum_{k=1}^{r} \binom{n}{k} \binom{m}{k} = \binom{n+m}{r} \]
where \( r \) is the number of toys exchanged.

### Step 6: Final Result
Thus, the total 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:
\[ \text{Total Ways} = \binom{n_1 + n_2}{k} \]
for \( k \) ranging from 1 to \( \min(n_1, n_2) \).

### Conclusion
The final answer can be expressed as:
\[ \text{Total Ways} = \sum_{k=1}^{\min(n_1, n_2)} \binom{n_1}{k} \cdot \binom{n_2}{k} \]