We will prove that:

\(\dbinom{n}{k} = \displaystyle \prod_{i=1}^{k} \frac{n-i+1}{i}\)

\(\forall\) \(k \in \mathbb{N}\), \(k \leq n\)

Proof #1: As the reader may guess, we will use induction to prove this. We establish a base case as \(k=1\):

\(\dbinom{n}{1} = \displaystyle \prod_{i=1}^{1} \frac{n-i+1}{i}\)

\(\Rightarrow\) \(\frac{n!}{1!(n-1)!} = n\)

\(\Rightarrow\) \(n=n\)

The base case clearly holds. Now we establish the inductive case as \(n=\alpha\):

\(\dbinom{n}{\alpha} = \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}\)

\(\Rightarrow\) \(\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\frac{n-\alpha}{\alpha +1} \displaystyle \prod_{i=1}^{\alpha} \frac{n-i+1}{i}\) (By hypothesis)

\(\Rightarrow\) \(\frac{n-\alpha}{\alpha +1} \dbinom{n}{\alpha} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}\)

\(\Rightarrow\) \(\frac{n!(n-\alpha)}{(\alpha +1)!(n-\alpha)!} =\displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}\)

\(\Rightarrow\) \(\frac{n!}{(\alpha +1)!(n-(\alpha+1))!} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}\)

\(\Rightarrow\) \(\dbinom{n}{\alpha+1} = \displaystyle \prod_{i=1}^{\alpha + 1} \frac{n-i+1}{i}\)

So we may conclude that the statement is true for \(k=1\) and if it is true for some \(k=\alpha\), then it must be true for \(k=\alpha+1\). The proof follows by induction.

QED

Proof #2: We will now offer an alternative proof, which is actually just a simple direct derivation:

\(\dbinom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{\displaystyle \prod_{i=0}^{n-1} (n-i)}{\displaystyle \prod_{i=0}^{k-1} (k-i) \left( \displaystyle \prod_{i=0}^{n-k-1} (n-k-i) \right)}\)

\( = \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{k!} = \frac{\displaystyle \prod_{i=1}^k (n-i+1)}{ \displaystyle \prod_{i=1}^k i }\)

\( = \displaystyle \prod_{i=1}^k \frac{(n-i+1)}{i} \) Because multiplication is commutative.

Which is the intended result.

QED.

## Comments

Sort by:

TopNewestI've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them! – Vikram Waradpande · 1 year, 9 months ago

Log in to reply

– Ethan Robinett · 1 year, 9 months ago

Thanks!Log in to reply

Nice note@Ethan Robinett – Anuj Shikarkhane · 2 years, 2 months ago

Log in to reply

– Ethan Robinett · 2 years, 2 months ago

Thanks appreciate it!Log in to reply

Hey nice proofs! :) I did the Inductive one but didn't think of using a direct one – Happy Melodies · 2 years, 2 months ago

Log in to reply

– Ethan Robinett · 2 years, 2 months ago

Thanks! Yeah after I did the induction, I figured there was probably a way to directly show it, given that factorials are just products by definition, I thought the second one was pretty interestingLog in to reply

I like the way the induction is used. – Akhil B Arackal · 2 years, 2 months ago

Log in to reply

– Ethan Robinett · 2 years, 2 months ago

Thanks manLog in to reply