6 Lecture 5, January 17, 2024

Properties of the Binomial coefficients

There are some useful/important results about permutation and combination.

  1. n(k)=n(n−1)(k−1) for k≥1
  2. {n \choose k} = \frac{n^{(k)}}{k!}
  3. {n \choose k} = {n \choose n-k} for k \geq 0
  4. {n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}
  5. Binomial theorem: (1 + x)^n = \sum_{k=0}^n {n \choose k} x^k
  6. {n \choose k} is equal to the kth entry in the nth row of Pascal’s triangle.

Note: Many of these idenetity may be proven using something called combinatorial proof. See Wiki for an (easy) example.

Proof (4). \begin{align*} {n-1 \choose k-1} + {n-1 \choose k} &= \frac{(n-1)!}{(k-1)! (n-k)!} + \frac{(n-1)!}{k! (n-k-1)!}\\ &= \frac{(n-1)!k }{(k-1)! (n-k)!k} + \frac{(n-1)!(n-k)}{k! (n-k-1)!(n-1)} \\ &= \frac{(n-1)!k + (n-1)! (n-k)}{k! (n-k)!} \\ &- \frac{(n-1)!( k + (n-k))}{k! (n-k)!} \\ &= \frac{(n-1)! n}{k! (n-k)!} \\ &= {n \choose k} \end{align*}

Aside: Stirling’s formula

n! grows really fast as n increases, so sometimes we need to approximate its value for computational reasons.

Stirling’s formula provides one such method, and it is given by

n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n, where \sim means their ratio approaches 1 as n goes to infinity.

We won’t need this approximation, but it’s useful to know it exists.

Example of use: Show that 2^{-2n}\binom{2n}{n} \approx \sqrt{\frac{2}{\pi n}}

6.1 Example in class

Example 6.1 (Application of Stirling's Formula/Approximation for factorial) Show that 2^{-2n}{2n \choose n} \approx \sqrt{\frac{1}{\pi n}}

\begin{align*} 2^{-2n}{2n \choose n} &= 2^{-2n}\frac{2n!}{n!n!} \\ &\approx 2^{-2n} \frac{\sqrt{2\pi (2n)} (2n/e)^{2n}}{\sqrt{2\pi n} (n/e)^{n}\sqrt{2\pi n} (n/e)^{n}}\\ &= 2^{-2n} \frac{\sqrt{4}}{\sqrt{2}\sqrt{2}} \frac{\sqrt{\pi n}}{\sqrt{\pi n}\sqrt{\pi n}} \frac{(2n)^{2n}}{n^n n^n} \frac{e^{-2n}}{e^{-2n}}\\ &= 2^{-2n} \frac{1}{\sqrt{\pi n}} 2^{2n}\\ &= \frac{1}{\sqrt{\pi n}} = \sqrt{\frac{1}{\pi n}} \end{align*}