How to prove $\binom{n+1}{m+1}=\binom{0}{m}+\binom{1}{m}+\dots+\binom{n}{m}$ combinatorially

Question

How can we prove combinatorially

$$\binom{n+1}{m+1}=\binom{0}{m}+\binom{1}{m}+\dots+\binom{n}{m}$$

I can get LHS by asking: How many ways can we form an $m+1$ person committee from a group of $n+1$ people. But I can't get RHS with this question.

I think I can get RHS by asking: How many ways can we form an $m$ person committee from a group of at most $n$ people. But I can't get LHS with this question.

possible duplicate of Prove ${n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}$ — Austin Mohr, Sep 30 '15 at 01:54
The formula is false (or meaningless) . It should be:$$\binom{n+1}{m+1}=\binom{m}{m}+\binom{m+1}{m}+\dots+\binom{n}{m}.$$ — Bernard, Sep 30 '15 at 01:54
If only had a penny for each time this question has been asked. — Asinomás, Sep 30 '15 at 02:01
On the other hand I've gained like 200 rep for this question over the years. — Asinomás, Sep 30 '15 at 02:02
@Bernard It is the same formula under the convention that $\binom{k}{m} = 0$ for $k < m$. — Austin Mohr, Sep 30 '15 at 02:08

score 3 · Accepted Answer · edited Sep 30 '15 at 02:04

3

Count how many ways to select $m+1$ people from a line of $n+1$ people, by selecting one person at some place (call it $k$), and then select $m$ people from the $k-1$ earlier in the line.

This count is $\sum\limits_{k=1}^{n+1} \binom{k-1}{m} = \sum\limits_{k=m+1}^{n+1}\binom{k-1}{m}$

edited Sep 30 '15 at 02:04

A.S.

4,004

answered Sep 30 '15 at 01:52

Graham Kemp

129,094

score 0 · Answer 2 · answered Sep 30 '15 at 01:56

0

Hint: Use Pascal's triangle identity $$ {n+1 \choose m+1} + {n+1 \choose m} = {n+2 \choose m+1} $$ and induction.

answered Sep 30 '15 at 01:56

William Stagner

4,786

How is this combinatorial? – user5826 Sep 30 '15 at 02:30
I suppose this is combinatorial in the sense that it would be acceptable in a combinatorics journal. – William Stagner Sep 30 '15 at 02:39
Usually combinatorial proof means double counting. – user5826 Sep 30 '15 at 02:45

How to prove $\binom{n+1}{m+1}=\binom{0}{m}+\binom{1}{m}+\dots+\binom{n}{m}$ combinatorially

2 Answers2