# Exercise C.1.11

$\star$ Argue that for any integers $n \ge 0, j \ge 0, k \ge 0$ and $j + k \le n$,

$$\binom{n}{j+k} \le \binom{n}{j}\binom{n-j}{k}$$

Provide both an algebraic proof and an argument based on a method for choosing $j + k$ items out of $n$. Give an example in which equality does not hold.

First, let's establish that, $j!k! \le (j+k)!$. Both sides have the same number of terms, but the right side has $k$ terms more - $(j+1)(j+2)\ldots(j+k)$ - that are greater than the corresponding terms on the left side $1\cdot2\cdot\ldots k$.

Thus:

$$\binom{n}{j} \binom{n-j}{k} = \frac{n!}{j!(n-j)!} \frac{(n-j)!}{k!(n-j-k)!} = \frac{n!}{j!k!(n-j-k)!} \ge \frac{n!}{(j+k)!(n-j-k)!} = \binom{n}{j+k}$$

As for the argument, the right side is the number of ways in which we can:

1. Choose $j$ elements out of $n$
2. Choose $k$ elements out of the remaining $n-j$ elements

There are more ways to do that than just choosing $j+k$ elements out of $n$, because this implies some ordering in the choice. For example, if $j = k = 1$, there are two ways to pick two elements (each one first) with this approach, but only one otherwise.

If $n = 4, j = k = 1$, then equality does not hold.