Discrete Mathematics - In how many ways can $A$ and $B$ exchange their books? $A$ has $5$ different books and $B$ has $3$ different books. How many ways are there in which they can exchange their books so that each keeps his individual number of books unchanged?
I got the answer as $55$, using combinations. Could you please tell me whether it is right, or point me in the right direction otherwise?
What if the question was the maximum possible ways in which they can be exchanged?? Is it P(5,3). Or Pow(5,3)..
 A: If $A$ and $B$ exchange $1$ book, there are ${5\choose 1}\cdot {3\choose 1} = 15$ ways they can do this. If they exchange $2$ books, then there are ${5\choose 2}\cdot {3\choose 2} = 30$ ways to do this, and if they exchange $3$ books, there are ${5\choose 3}\cdot {3\choose 3} = 10$. All together, there are $15+30+10=55$ ways to do it. Your answer is correct.
A: Another approach:
After any exchange, $A$ has to have $5$ books, and $B$ has to have (the remaining) $3$ books. So the result of an exchange is exactly the same as the result of selecting $5$ of the $8$ distinct books and giving them to $A$, and giving the remaining $3$ to $B$.
The number of ways of selecting $5$ books from $8$ distinct books is $\binom 8 5 = 56$.
Since they already have one such selection of books, the number of ways they can exchange them to get other selections is $56 - 1 = 55$.
Alternative Interpretation:
The question could be asking about the number of possible exchanges, where an exchange is defined as $\{A_i \leftrightarrow B_1, A_j \leftrightarrow B_2, A_k \leftrightarrow B_3 \}$, where $A_i$s and $B_p$s are the books currently with $A$ and $B$ (listed in some order). Then, as Deuteu was saying, $\{A_1 \leftrightarrow B_1, A_2 \leftrightarrow B_2, A_3 \leftrightarrow B_3\}$ is different from $\{A_1 \leftrightarrow B_2, A_2 \leftrightarrow B_1, A_3 \leftrightarrow B_3\}$.
The number of such exchanges is, of course, the number of ways of picking $A_i, A_j, A_k$ from $A_1, \ldots, A_5$, with understanding that $A_i$ is exchanged with $B_1$, then $A_j$ with $B_2$, and $A_k$ with $B_3$ (so the order matters). The answer is, therefore, $^5\text{P}_3 = 60$.
