More about k-subsets 10/9

Summary of taking k elements out of n:

	Order matters	Order does not matter
With repetitions	nk	$\binom{n+k-1}{n-1}=\binom{n+k-1}{k}$
Without repetitions	$n\times (n-1) \times \ldots (n-k+1)=n_k$	$\binom{n}{k}$

Because the number of all subsets of an n-set, whatever their cardinal is 2ⁿ, we have :

$$\binom{2n}{0}+ \binom{2n}{1} \binom{2n}{2} + \cdots +\binom{2n}{2n}=2^{2n}$$

Thus

$$\binom{2n}{n} \leq 2^{2n}$$

On the other hand each term of this sum is smaller than the largest one, which is:

$$\binom{2n}{n}$$

which is thus larger than their average:

$$\frac{2^{2n}}{2n+1} \leq \binom{2n}{n}$$

Another way of considering the numbers of ways of choosing k sets out of n sets, with repetitions allowed is the number of ways of dropping k balls into n boxes, this image is an image often used in statistical mechanics. As I pointed out in class, there are several ways of putting probabilities on these spaces.