Can someone pls show that
nCr = n(n-1)(n-2) ... (n-r+1)
r(r-1)*...*2*1
$\displaystyle _nC_r = \frac{n!}{r!(n-r)!} = \frac{n(n-1)..(n-r+1)(n-r)..(1)}{r!(n-r)!}$$\displaystyle =\frac{n(n-1)..(n-r+1)r!}{r!(n-r)!}=\frac{n(n-1)..(n-r+1)}{r!}$
or do you mean something else?
Choosing $\displaystyle r$ from $\displaystyle n$ distinct elements, the first may be choosen in $\displaystyle n$ ways the second in $\displaystyle (n-1)$ down to the $\displaystyle r$-the in $\displaystyle (n-r+1)$ ways, so there are $\displaystyle n(n-1)..(n-r+1)$ ways of doing the selection, but as each combination appears in these choices in ever permutation we are counting each combination r! times so the number of combinations is:
$\displaystyle \frac{n(n-1)..(n-r+1)}{r!}$
RonL