We know that
$$\dbinom{n}r = \dfrac{n!}{(n-r)!r!}$$
An intuitive explanation of the formula is that, if I partition the total number of permutations of objects by $r!$, and choose one member of each partition, then no similarly ordered pattern will be registered more than once.
Is there a more intuitive explanation than this?
Best Answer
Think of it this way. Suppose that I have balls numbered $1$ through $n$, I want to pick $r$ of them, and I do it one ball at a time. There are $n$ different ways in which I could choose the first ball. Once it’s been chosen, there are only $n-1$ balls left, so there are $n-1$ different ways in which I could pick the second ball. Thus, there are $n(n-1)$ different ways in which I could pick the first two balls. If I continue in this fashion, after I’ve picked $r-1$ balls, there will be $n-(r-1)=n-r+1$ balls left, so I’ll be able to pick the $r$-th ball in $n-r+1$ different ways. Thus, the sequence of $r$ balls can be chosen in
$$n(n-1)(n-2)\ldots(n-r+1)=\frac{n!}{(n-r)!}\tag{1}$$
different ways. The $(n-r)!$ in the denominator merely serves to cancel out the unwanted factors in $n!$.
However, $(1)$ is the number of different sequences of $r$ balls that we could choose: if $B$ is a set of $r$ balls, $(1)$ counts every possible permutation of the balls in $B$ separately. For example, if $B=\{b_1,b_2,b_3\}$, expression $(1)$ counts $B$ six times, once for each of the $3!=6$ possible sequences in which we could have chosen $B$: $\langle b_1,b_2,b_3\rangle,\langle b_1,b_3,b_2\rangle,\langle b_2,b_1,b_3\rangle,\langle b_2,b_3,b_1\rangle,\langle b_3,b_1,b_2\rangle$, and $\langle b_3,b_2,b_1\rangle$.
The same reasoning that I used to arrive at $(1)$ shows that there are $r!$ different ways to arrange a set of $r$ balls in order, so each $r$-element set of balls has been counted $r!$ times in $(1)$. Thus, to get the actual number of $r$-element subsets of our set of $n$ balls, we must divide $(1)$ by $r!$, getting
$$\binom{n}r=\frac{n!}{(n-r)!r!}\;$$
From the standpoint of seeing why this works, it’s better to think of it as
$$\frac{n(n-1)(n-2)\ldots(n-r+1)}{r!}=\frac{\prod_{i=0}^{r-1}(n-i)}{r!}\;;$$
that actually reflects the reasoning involved.