Question

10 people are sitting around a circular table, each one shaking a hand with everyone else except from the people sitting on either side of him. Find the number of handshakes.

Answer 1

35

Answer 2

Let $N$ be the number of people sitting around a circular table. Here, $N = 10$.

The total number of handshakes if every person shakes hands with every other person is given by the number of ways to choose 2 people out of $N$, which is $\binom{N}{2}$.

Total possible handshakes = $\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$.

The problem states that people sitting on either side of a person do not shake hands. This means that adjacent people around the table do not shake hands.

In a circular arrangement of 10 people, the adjacent pairs are $(P_1, P_2), (P_2, P_3), \dots, (P_9, P_{10}), (P_{10}, P_1)$.

There are exactly $N=10$ such adjacent pairs.

These 10 pairs represent the handshakes that are forbidden by the rule.

The number of handshakes that actually occur is the total number of possible handshakes minus the number of forbidden handshakes.

Number of handshakes = (Total possible handshakes) - (Number of forbidden handshakes)

Number of handshakes = $\binom{10}{2} - 10$

Number of handshakes = $45 - 10 = 35$.

Alternatively, consider a single person. This person does not shake hands with themselves (1 person) and does not shake hands with their two neighbours (2 people). So, each person shakes hands with $N - 1 - 2 = 10 - 1 - 2 = 7$ people.

Since there are 10 people, the total count of handshake "ends" is $10 \times 7 = 70$.

As each handshake involves two people, the total number of handshakes is half of this count.

Number of handshakes = $\frac{10 \times 7}{2} = \frac{70}{2} = 35$.

Both methods yield the same result.