Waste less time on Facebook — follow Brilliant.
×

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, 3 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Vikram Waradpande - 2 years, 10 months ago

Log in to reply

Thanks!

Ethan Robinett - 2 years, 10 months ago

Log in to reply

Nice note@Ethan Robinett

Anuj Shikarkhane - 3 years, 3 months ago

Log in to reply

Thanks appreciate it!

Ethan Robinett - 3 years, 3 months ago

Log in to reply

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

Happy Melodies - 3 years, 3 months ago

Log in to reply

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

Ethan Robinett - 3 years, 3 months ago

Log in to reply

I like the way the induction is used.

Akhil B Arackal - 3 years, 3 months ago

Log in to reply

Thanks man

Ethan Robinett - 3 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...