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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 :

\begin{displaymath}\binom{2n}{0}+ \binom{2n}{1}
\binom{2n}{2} + \cdots +\binom{2n}{2n}=2^{2n}\end{displaymath}

Thus

\begin{displaymath}\binom{2n}{n} \leq 2^{2n}\end{displaymath}

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

\begin{displaymath}\binom{2n}{n}\end{displaymath}

which is thus larger than their average:

\begin{displaymath}\frac{2^{2n}}{2n+1} \leq \binom{2n}{n} \end{displaymath}

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.



Susan Holmes
1998-12-07