Question

If n people are sitting on a circle, then what are the number of ways in which 3 people can be selected such that no two of them are consecutive.

• A. n(n-4)(n-5)/6
• B. n(n-5)(n-6)/6
• C. n(n-3)(n-4)/6
• D. (n-1)(n-2)(n-3)/6

Answer 1

Answer: (A)

n(n-4)(n-5)/6

Answer 2

Let $n$ be the number of people sitting in a circle. We need to select 3 people such that no two of them are consecutive.

Let the positions of the three selected people be $p_1, p_2, p_3$ in increasing order, where $1 \le p_1 < p_2 < p_3 \le n$. For no two people to be consecutive, the difference between their positions must be at least 2. So, we must have:

1. $p_2 - p_1 \ge 2$
2. $p_3 - p_2 \ge 2$
3. $p_1 + n - p_3 \ge 2$ (This accounts for the circular arrangement).

Let's define new variables: $x_1 = p_2 - p_1$ $x_2 = p_3 - p_2$ $x_3 = n - p_3 + p_1$

Then $x_1 + x_2 + x_3 = (p_2 - p_1) + (p_3 - p_2) + (n - p_3 + p_1) = n$. Since no two people are consecutive, $x_1 \ge 2$, $x_2 \ge 2$, $x_3 \ge 2$.

Let $y_1 = x_1 - 2$, $y_2 = x_2 - 2$, $y_3 = x_3 - 2$. Then $y_1, y_2, y_3 \ge 0$. Substituting: $(y_1+2) + (y_2+2) + (y_3+2) = n$ $y_1 + y_2 + y_3 + 6 = n$ $y_1 + y_2 + y_3 = n-6$.

The number of non-negative integer solutions is $\binom{(n-6) + 3 - 1}{3 - 1} = \binom{n-4}{2}$.

This is equivalent to the formula $\frac{n}{k} \binom{n-k-1}{k-1}$ for $k=3$: Number of ways = $\frac{n}{3} \binom{n-3-1}{3-1} = \frac{n}{3} \binom{n-4}{2}$.

Expanding: $\frac{n}{3} \binom{n-4}{2} = \frac{n}{3} \frac{(n-4)!}{2!(n-6)!} = \frac{n}{3} \frac{(n-4)(n-5)}{2} = \frac{n(n-4)(n-5)}{6}$.

We must have $n-4 \ge 2$, which means $n \ge 6$. If $n < 6$, the number of ways is 0.

Therefore, the final answer is $\frac{n(n-4)(n-5)}{6}$.