Let $0 \leq r \leq n$. Can someone give an intuitive explanation for why the following identity holds or give me a hint on proving it?
$\dbinom{n}{n-r} = \dbinom{n}{r}$
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
One way to pick $3$ among $10$ apples, is to pick the $7$ that you will not keep.
answered Feb 5, 2012 at 1:30
$\begingroup$
$\endgroup$
$\binom{n}r$ is the number of ways to choose $r$ objects from a set of $n$ objects. When you choose $r$ objects, you’re rejecting the other $n-r$ objects, so $\binom{n}r$ is also the number of ways to pick $n-r$ of the $n$ objects to be rejected. But that’s clearly the same as the number of ways to choose $n-r$ of the $n$ objects, which is $\binom{n-r}r$: choosing in order to keep and choosing in order to reject are both just choosing.
answered Feb 5, 2012 at 1:32