(a) There are $n$ chairs in a row. Find the number of ways of choosing $k$ of these chairs, so that no two chosen chairs are adjacent. (b) There are 10 chairs in a circle, labelled from 1 to 10. Find the number of ways of choosing 3 of these chairs, so that no two chosen chairs are adjacent.

(c) There are $n$ chairs in a circle, labelled from 1 to $n.$ Find the number of ways of choosing $k$ of these chairs, so that no two chosen chairs are adjacent.

(A) P (n,k) = n!/(n-k)! divided by 2

(B) C (n,3) = n!/(12)(n-3)!

(C) C (n,k) = n!/(n-k)!(k!)

Step-by-step explanation:

Permutation deals with order or arrangement or position of objects. Where this does not matter, we use the Combination formula.

We divide by 2 in all cases, because no 2 chosen chairs should be adjacent.

For (B), n=10

C (n,3) = n!/(n-3)!(3!) divided by 2

3! = 3×2×1 = 6

The expression divided by 2 means it will be multiplied by 1/2

Hence 6×2 = 12

And we arrive at

C (n,3) =n!/(12)(n-3)!

idk this answer sorry cant on this one.

step-by-step explanation: