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 \(k \geq 1\)
  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 \(k\)th entry in the \(n\)th 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 \(2^{-2n}{2n \choose n} \approx \sqrt{\frac{1}{\pi n}}\)

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