# Combination Proof

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.

Note by Ethan Robinett
3 years, 10 months ago

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I've been scrolling through your profile the complete morning now and am finding some really nice notes. Thanks for sharing them!

- 3 years, 4 months ago

Thanks!

- 3 years, 4 months ago

Nice note@Ethan Robinett

- 3 years, 10 months ago

Thanks appreciate it!

- 3 years, 10 months ago

Hey nice proofs! :) I did the Inductive one but didn't think of using a direct one

- 3 years, 10 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 interesting

- 3 years, 10 months ago

I like the way the induction is used.

- 3 years, 10 months ago

Thanks man

- 3 years, 10 months ago