What's the combinational interpretation of the identity $\sum\limits_{k=r}^{n} {k \choose r }={{n+1} \choose r+1}$?
Asked
Active
Viewed 279 times
0
-
Just use induction and the identity: $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.$$ – Jack D'Aurizio Nov 08 '14 at 18:58
-
I know how to prove it by induction but I need to undestand it's combinatorial interpretation. – Breldor Nov 08 '14 at 19:00
-
4Then count the number of subsets of ${1,\ldots,n+1}$ with $r+1$ elements according to the maximum element in them. – Jack D'Aurizio Nov 08 '14 at 19:02
-
More details are given in http://math.stackexchange.com/a/357087/139123 – David K Nov 08 '14 at 21:04
1 Answers
2
we want choose r+1 persons from n+1 persons let A(1), A(2),..., A(n+1) are that n+1 persons first we count the numbers of choices that A(1) is choosed that equals to (r,n) second we count the numbers of choices that A(1) is not choosed but A(2) is choosed that equals to (r,n-1) third... and sum of these is equal to (r+1,n+1)
GhD
- 926