On a possible generalization of Pascal's formula

Question

I have observed the following relation

$$\binom{n}{k}=\sum\limits_{i=0}^{k} \binom{n-i-1}{k-i}, \quad \frac{n}2\leq k \leq n-1$$

Numerical example with $n=5$ and $k=3$

$$\binom{\color{red}5}{\color{blueviolet}3}= \binom{\color{Magenta}4}{\color{blueviolet}3}+\binom{\color{blueviolet}3}{\color{Orange}2}+\binom{\color{Orange}2}{\color{YellowGreen}1}+\binom{\color{YellowGreen}1}{\color{Brown}0}$$

$$10\ \ \ = \ \ \ 4 \ \ \ +\ \ \ 3 \ \ \ + \ \ \ 2 \ \ \ \ + \ \ 1$$

Is this relation true? If yes, is there a name for it?

I´ve used identities like $\binom{n+1}{k+1} = \binom nk + \binom n{k+1}$ to show the relation above, but I failed. Does anyone have an idea how to manage it?

Thanks for taking time to read the question.

It is an instance of the hockeystick identity. For a proof of it see here. — drhab, Dec 08 '19 at 10:38

score 7 · Accepted Answer · answered Dec 08 '19 at 10:50

7

See my comment on your question.

The hockeystick identity tells us that: $$\sum_{i=r}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$

Applying this we find:

$$\sum_{i=0}^{k}\binom{n-i-1}{k-i}=\sum_{i=n-1-k}^{n-1}\binom{i}{n-1-k}=\binom{n}{n-k}=\binom{n}{k}$$

answered Dec 08 '19 at 10:50

drhab

153,781

score 6 · Answer 2 · answered Dec 08 '19 at 11:52

6

This can also be shown with negative binomial coefficients and Vandermonde's Identity: $$ \begin{align} \sum_{i=0}^k\binom{n-i-1}{k-i} &=\sum_i\binom{n-i-1}{k-i}\binom{i}{i}\tag1\\ &=\sum_i\binom{k-n}{k-i}(-1)^{k-i}\binom{-1}{i}(-1)^i\tag2\\ &=(-1)^k\binom{k-n-1}{k}\tag3\\[3pt] &=\binom{n}{k}\tag4 \end{align} $$ Explanation:
$(1)$: remove the limits from the summation
$(2)$: negative binomial coefficients
$(3)$: Vandermonde Identity
$(4)$: negative binomial coefficients

answered Dec 08 '19 at 11:52

robjohn

353,833

1

Finally I found that $\binom{i}{i}=\binom{-1}{i}(-1)^i$ also uses negative binomial coefficients. I always learn a lot from your answer! – linear_combinatori_probabi Sep 05 '20 at 09:26
2

@FtyRain: Indeed. In $(7)$ of this answer, I state $\binom{-1}{i}=(-1)^i[i\ge0]$, and of course, $\binom{i}{i}=[i\ge0]$. – robjohn Sep 05 '20 at 09:57

score 1 · Answer 3 · answered Mar 04 '25 at 05:47

I´ve used identities like $\binom{n+1}{k+1} = \binom nk + \binom n{k+1}$ to show the relation above, but I failed.

Each pair of square brackets below denotes an application of Pascal's formula $\binom{n+1}{k+1} = \binom nk + \binom n{k+1}$:

\begin{align*} \binom{n}{k}&=\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\\ &=\binom{n-1}{k}+\left[\binom{n-2}{k-1}+\binom{n-2}{k-2}\right]\\ &=\binom{n-1}{k}+\binom{n-2}{k-1}+\left[\binom{n-3}{k-2}+\binom{n-3}{k-3}\right]\\ &\cdots\\ &=\binom{n-1}{k}+\binom{n-2}{k-1}+\binom{n-3}{k-2}+\cdots+\left[\binom{n-k}{k-(k-1)}+\binom{n-k}{k-k}\right]\\ &=\binom{n-1}{k}+\binom{n-2}{k-1}+\binom{n-3}{k-2}+\cdots+\binom{n-k}{k-(k-1)}+\binom{n-k-1}{k-k}\\ &=\sum_{i=0}^k\binom{n-i-1}{k-i}.\\ \end{align*}

On a possible generalization of Pascal's formula

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