Six friends will exchange books in their book club. Each friend has one book to give to a friend, and will receive one book from a different friend. (No two friends trade books with each other.) In how many ways can the books be exchanged?
(A) 200
(B) 120
(C) 140
(D) 240
(E) 160
We solve this problem by first determining the possible structures of the exchange within the book club. Then, we count the number of ways the friends can be fit into the structures.

Suppose that the six friends exchange books as described.

Consider one person, who we call $A$.

$A$ must give a book to a second person, who we call $B(A \rightarrow B)$.

$B$ cannot give a book back to $A$, so must give a book to a third person, who we call $C$ $(A \rightarrow B \rightarrow C)$.

$C$ cannot give a book to $B$ since $B$ already receives a book from $A$. Therefore, $C$ can give a book to $A(A \rightarrow B \rightarrow C \rightarrow A)$ or $C$ can give a book to a fourth person, who we call $D$ $(A \rightarrow B \rightarrow C \rightarrow D)$.

- In the first case $(A \rightarrow B \rightarrow C \rightarrow A$ ), each of $A, B$ and $C$ already both give and receive a book. Therefore, the fourth person, who we call $D$, must give a book to a fifth person, who we call $E(A \rightarrow B \rightarrow C \rightarrow A ; D \rightarrow E)$.

$E$ cannot give a book back to $D$ and cannot give to any of $A, B$ or $C$, so must give a book to the sixth (and final) person, who we call $F(A \rightarrow B \rightarrow C \rightarrow A ; D \rightarrow E \rightarrow F)$.

The only person that $F$ can give a book to is $D$, since everyone else already is receiving a book. This completes this case as $(A \rightarrow B \rightarrow C \rightarrow A ; D \rightarrow E \rightarrow F \rightarrow D)$.

- The other option from above is $A \rightarrow B \rightarrow C \rightarrow D$.

Here, $D$ cannot give a book to $B$ or $C$ who each already receive a book. Thus, $D$ gives a book to $A$ or to one of the two remaining people.

If $D$ gives a book to $A$, then each of $A, B, C$ and $D$ both gives and receives, so the final two people are left to exchange books with each other, which is not possible.

Therefore, $D$ gives a book to one of the remaining people, who we call $E$ (giving $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E)$.

Only two people have not received a book now: $A$ and the sixth person, who we call $F$. $E$ must give a book to $F$, otherwise $F$ does not give or receive a book at all. Thus, $F$ must give a book to $A$.

This gives the structure $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow A$.

Therefore, there are two possible structures for the exchanges: $A \rightarrow B \rightarrow C \rightarrow A$ with $D \rightarrow E \rightarrow F \rightarrow D$, and $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow A$.

(We can think of the first structure as two cycles of three people, and the second as one cycle of six people.)

Suppose the six friends are $P, Q, R, S, T$, and $U$. (We could call them Peter, Quinn, Rad, Steve, Troy, and Ursula, but will use abbreviations to save space.) We must count the number of ways that these six friends can be assigned to the positions $A$ through $F$ in the two structures above. We have to be careful because assigning

$$
\begin{aligned}
& A \rightarrow B=C \\
& P \\
& Q
\end{aligned}
$$

in the first structure is the same as assigning

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-138.jpg?height=120&width=1321&top_left_y=2363&top_left_x=462)

but is different from assigning

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-139.jpg?height=112&width=1321&top_left_y=150&top_left_x=462)

- Case 1: $A \rightarrow B \rightarrow C \rightarrow A$ with $D \rightarrow E \rightarrow F \rightarrow D$

Because this case consists of two cycles of three people each, each person serves exactly the same function, so we can assign $P$ to position $A$.

There are then 5 possible friends for position $B$ (all but $P$ ).

For each of these, there are 4 possible friends for position $C$ (all put $P$ and the friend in position $B$ ). This completes the first cycle.

Now, the remaining three friends must complete the second cycle.

Suppose without loss of generality that these three friends are $S, T$ and $U$.

Since each of the three positions in the second cycle serves exactly the same function, we can assign $S$ to position $D$.

There are then 2 possible friends for position $E$ (either $T$ or $U$ ).

For each of these, there is 1 possible friend left to go in position $F$.

This gives $5 \times 4 \times 2 \times 1=40$ ways of assigning the friends to this structure.

- Case 2: $A \rightarrow B \rightarrow C \rightarrow D \rightarrow E \rightarrow F \rightarrow A$

Because this case is a single cycle of six people, each person serves exactly the same function, so we can assign $P$ to position $A$.

There are then 5 possible friends for position $B$ (all but $P$ ).

For each of these, there are 4 possible friends for position $C$ (all put $P$ and the friend in position $B$ ).

Similarly, there are 3 possibilities for position $D, 2$ for position $E$, and 1 for position $F$. This gives $5 \times 4 \times 3 \times 2 \times 1=120$ ways of assigning the friends to this structure.

Therefore, there are $40+120=160$ ways in total in which the books can be exchanged.

ANSWER: (E)