To explain: why the combination binom{n}{r}=binom{n}{r,n-r}.

Question

To explain: why the combination $(\binom{n}{r}) = (\binom{n}{r, n - r})$

Answer 1

Formula used:
The number of combinations of choosing r objects from n objects is
$^{{n}} C_{r} = \frac{n!}{(n - r) r!}$
The number of combinations of choosing two groups x and y from n objects is
$^{{n}} C_{x, y} = \frac{n!}{x! y! (n - x - y)!}$
Consider the number of combinations of choosing r objects from n objects $= (\binom{n}{r})$
Let n is divided in to two groups one is having r objects and other group is having n
$(n - r)$ Objects, that is $n = r + (n - r)$ , then
The number of ways of Choosing two groups r and $n - r$ from n objects $= (\binom{n}{r, n - r})$
$(\binom{n}{r, n - r}) = \frac{n!}{r! (n - r)! (n - (n - r) - r)!}$
$= \frac{n!}{r! (n - r) (n - n + r - r)}$
$= \frac{n!}{r! (n - r)!}$
$= (\binom{n}{r})$
Therefore, $(\binom{n}{r}) = (\binom{n}{r, n - r})$
Hence, explained

To explain: why the combination binom{n}{r}=binom{n}{r,n-r}.

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