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As I understand Pascal's identity is defined as

$$\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$$

Just as an exercise I decided to prove this identity. Here is my solution:

$$ \begin{align*} \binom{n - 1}{k - 1} + \binom{n - 1}{k} = \dfrac{(n-1)!}{(k-1)!(n-1-(k-1))!} + \dfrac{(n-1)!}{k!(n - 1 - k)!} = \\ \dfrac{(n-1)!}{(k-1)!(n-1-k+1)!} + \dfrac{(n-1)!}{k!(n-(k+1))!} = \\ \dfrac{(n-1)!}{(k-1)!(n-k)!} + \dfrac{(n-1)!}{k!(n - (k+1))!} = \\ \dfrac{(n - 1)!}{(k-1)!(n-k)(n-(k+1))!} + \dfrac{(n-1)!}{k(k-1)!(n-(k+1))!} = \\ \dfrac{k(n-1)! + (n-k)(n-1)!}{k(k-1)!(n-k)(n-(k+1))!} = \dfrac{(n-1)!(k+n-k)}{k!(n-k)!} = \\ \dfrac{n(n-1)!}{k!(n-k)!} = \dfrac{n!}{k!(n-k)!} = \binom{n}{k} \end{align*} $$ Given that I've just started learning about this identity I just wanted to make sure that my solution is correct. Is it right?

Robert Shore
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Odyssey2001
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    That is fine for a basic algebraic proof but it really should have been done by induction. A combinatorial proof however is far cleaner and easier. The subsets of size $k$ from the set ${1,2,\dots,n}$ can be partitioned into those who do include $n$ itself and those who do not. – JMoravitz Apr 11 '22 at 21:49

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Yes, it's perfectly fine. Maybe a little long-winded/verbose: the notation simplifies considerably if/when you pull out the common factors in numerator and denominator, before combining the fractions.

For that matter, if you've shown that the binomial coefficients are the (aptly-named) coefficients of $(x+y)^n$, then this "combinatorial identity" follows from $(x+y)^n=(x+y)^{n-1}\cdot (x+y)$ and the (provable) "rules" for multiplying polynomials. :)

The latter argument seems much more robust than any formulaic argument. (Current labels about "algebraic" versus "combinatorial" have unclear connotations for me...) In particular, I can see that $(x+y)^{n-1}\cdot (x+y)^n$ is true, and that it will entail identities among coefficients. Then it's not hard to see what those identities are... In contrast, just looking at "bare" binomial coefficients, I would certainly not guess that identity... though, once told that it was so, could have hope to prove it. But I want means to discover true things, not merely verify assertions from an oracle. :)

paul garrett
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