Five monkeys are seated around a table. Their seats are labelled $P, Q, R, S$, and $T$, in clockwise order, as shown. The five monkeys are randomly numbered Monkey 1, Monkey 2, Monkey 3, Monkey 4, and Monkey 5. Monkey 1 remains in its seat. The remaining four monkeys then sit themselves in the remaining seats so that they are seated in clockwise order as Monkey 1, Monkey 2, Monkey 3, Monkey 4, and Monkey 5. What is the probability that the Monkey originally in seat $R$ moves to seat $P$ ?
(A) $\frac{1}{20}$
(B) $\frac{1}{10}$
(C) $\frac{3}{20}$
(D) $\frac{1}{5}$
(E) $\frac{1}{4}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-071.jpg?height=293&width=307&top_left_y=257&top_left_x=1386)
Since the five monkeys are randomly numbered, then the probability that any given monkey is numbered Monkey 1 is $\frac{1}{5}$. There are five possibilities to consider.

- Case 1: If the monkey in seat $P$ is numbered Monkey 1, then it stays in its seat, so the monkey in seat $R$ cannot move to seat $P$.
- Case 2: If the monkey in seat $Q$ is numbered Monkey 1, then after the monkeys have moved, Monkey 2 will be in seat $R$, Monkey 3 in seat $S$, Monkey 4 in seat $T$, and Monkey 5 in seat $P$. Thus, if Monkey 1 is in seat $Q$, then the monkey in seat $R$ moves to seat $P$ if it was numbered Monkey 5.
- Case 3: If the monkey in seat $R$ is numbered Monkey 1, then it stays in its seat and so cannot move to seat $P$.
- Case 4: If the monkey in seat $S$ is numbered Monkey 1, then after the monkeys have moved, Monkey 2 will be in seat $T$, Monkey 3 in seat $P$, Monkey 4 in seat $Q$, and Monkey 5 in seat $R$. Thus, if Monkey 1 is in seat $S$, then the monkey in seat $R$ moves to seat $P$ if it was numbered Monkey 3.
- Case 5: If the monkey in seat $T$ is numbered Monkey 1, then after the monkeys have moved, Monkey 2 will be in seat $P$, Monkey 3 in seat $Q$, Monkey 4 in seat $R$, and Monkey 5 in seat $S$. Thus, if Monkey 1 is in seat $T$, then the monkey in seat $R$ moves to seat $P$ if it was numbered Monkey 2.

The possible cases that we have to consider are Cases 2, 4 and 5 .

- From Case 2, what is the probability that the monkey in seat $Q$ is numbered Monkey 1 and the monkey in seat $R$ is numbered Monkey 5 ? We know that the probability that the monkey in seat $Q$ is numbered Monkey 1 is $\frac{1}{5}$. Given this and the fact that the remaining four monkeys are numbered randomly, the probability is $\frac{1}{4}$ that the monkey in seat $R$ is numbered Monkey 5. Therefore, the probability of this combined event happening is $\frac{1}{5} \cdot \frac{1}{4}=\frac{1}{20}$.
- Similarly, from Case 4 the probability that the monkey in seat $S$ is numbered Monkey 1 and the monkey in seat $R$ is numbered Monkey 3 is also $\frac{1}{20}$.
- Also, from Case 5 the probability that the monkey in seat $T$ is numbered Monkey 1 and the monkey in seat $R$ is numbered Monkey 2 is $\frac{1}{20}$.

Therefore, the probability that the monkey who was in seat $R$ moves to seat $P$ is $3 \cdot \frac{1}{20}=\frac{3}{20}$.

ANSWER: (C)