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I tried to solve a question in combinatorics and ended up with this sum:

$\sum_{t=0}^{r}\binom{ m-t}{r-t}$

when I give it to WolframAlpha it says it equals to $\binom{ m + 1}{r}$ but gives no proof, can someone provide a discrete proof?

Ofer Magen
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2 Answers2

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Hint:

How many $r$-element subsets are there of $\{0,1,2,\dots,m\}$?

How many of them don't contain $0$?

How many contain $0$ but not $1$?

How many contain $0$ and $1$ but not $2$?

...

How many contain the first $t$ many numbers but not the $t+1$'st?

JMoravitz
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  • any discrete way? perhaps using induction? – Ofer Magen Dec 02 '17 at 17:08
  • @OferMagen what is unsatisfactory about a combinatorial proof? Most people prefer this method of proof as it gives more intuition as to what is going on. This could easily be reworded with symbols if you absolutely insist, by writing $\binom{[n+1]}{r}=E_0\sqcup E_1\sqcup\dots \sqcup E_r$ where $E_t={{0,1,2,\dots,t-1}\cup A~:~A\in\binom{{t+1,t+2,\dots,m}}{r-t}}$, but that just muddies things up in my opinion. – JMoravitz Dec 02 '17 at 17:23
  • because when you have a combinatorial question and you use a completely different combinatorial question the proof gets longer – Ofer Magen Dec 02 '17 at 17:24
  • plus induction is more "set in stone" than logic – Ofer Magen Dec 02 '17 at 17:25
  • what is a r-element subset – Ofer Magen Dec 02 '17 at 17:26
  • @OferMagen, the answer to your original question, to prove that $\sum\limits_{t=0}^r\binom{m-t}{r-t}=\binom{m+1}{r}$, can be answered with as few words as "The right hand side counts the number of $r$-element subsets of ${0,1,2,\dots,m+1}$ while the left hand side counts the same by breaking into cases based on the number of consecutive numbers starting at zero are in the subset." This requires no fancy algebra, no fancy computations, and hardly any work at all beyond recognizing what each term in the sum represents and recognizing why they both represent the same total. – JMoravitz Dec 02 '17 at 17:27
  • "What is an $r$-element subset" it is a subset with $r$ elements... – JMoravitz Dec 02 '17 at 17:28
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$\sum_{t=0}^r \binom{m-t}{r-t} = \sum_{t=0}^r \binom{m-t}{m - r} = \sum_{t=m-r}^m \binom{t}{m - r} = \binom{m + 1}{m-r + 1} = \binom{m + 1}{r}$

the third $= $ is the Hockey-stick identity https://en.wikipedia.org/wiki/Hockey-stick_identity

Ofer Magen
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