Question

Number of ways when r people are selected out of n people such that they are not consecutive in a circle are

Answer 1

$\frac{n}{n-r} \binom{n-r}{r}$

Answer 2

To find the number of ways to select $r$ people out of $n$ people arranged in a circle such that no two selected people are consecutive, we can use a standard combinatorial approach.

Let $N$ be the total number of ways. We consider a specific person, say $P_1$. There are two cases:

Case 1: $P_1$ is selected.

If $P_1$ is selected, then its immediate neighbors, $P_n$ and $P_2$, cannot be selected. This leaves us with $n-3$ people ($P_3, P_4, \dots, P_{n-1}$) arranged in a line. From these $n-3$ people, we need to select $r-1$ people such that no two are consecutive.

The number of ways to select $k$ non-consecutive items from $m$ items arranged in a line is given by $\binom{m-k+1}{k}$.

In this case, $m = n-3$ and $k = r-1$.

So, the number of ways for Case 1 is $\binom{(n-3)-(r-1)+1}{r-1} = \binom{n-3-r+1+1}{r-1} = \binom{n-r-1}{r-1}$.

Case 2: $P_1$ is not selected.

If $P_1$ is not selected, then we need to select $r$ people from the remaining $n-1$ people ($P_2, P_3, \dots, P_n$) arranged in a line such that no two are consecutive.

In this case, $m = n-1$ and $k = r$.

So, the number of ways for Case 2 is $\binom{(n-1)-r+1}{r} = \binom{n-r}{r}$.

The total number of ways is the sum of the ways from Case 1 and Case 2:

$N = \binom{n-r-1}{r-1} + \binom{n-r}{r}$.

We can simplify this expression. Let's use the identity for simplification: $\binom{n-r-1}{r-1} + \binom{n-r}{r}$ $= \frac{(n-r-1)!}{(r-1)!(n-2r)!} + \frac{(n-r)!}{r!(n-2r)!}$

Factor out common terms: $= \frac{(n-r-1)!}{(r-1)!(n-2r)!} \left( 1 + \frac{n-r}{r} \right)$ $= \frac{(n-r-1)!}{(r-1)!(n-2r)!} \left( \frac{r + n-r}{r} \right)$ $= \frac{(n-r-1)!}{(r-1)!(n-2r)!} \left( \frac{n}{r} \right)$

Rearrange terms to form a binomial coefficient: $= \frac{n}{r} \frac{(n-r-1)!}{(r-1)!(n-2r)!}$

To get $\binom{n-r}{r}$, we need $\frac{(n-r)!}{r!(n-2r)!}$. We have $(n-r-1)!$ in the numerator and $(r-1)!$ in the denominator.

$= \frac{n}{n-r} \frac{(n-r)!}{r!(n-2r)!}$ $= \frac{n}{n-r} \binom{n-r}{r}$

This formula is valid for $n \ge 2r$. If $n < 2r$, it's impossible to select $r$ non-consecutive people.