How do I find the number of subsets of a specific cardinality Given a set S with cardinality N, does a formula exist to calculate the number of subsets  with cardinality M?
For instance, if N=5, and M=3; how many subsets are there within S with cardinality M?
Here is some extra info for each cardinality...maybe it helps.

Cardinality 0 = 1 subset
Cardinality 1 = 5 subsets
Cardinality 2 = 10 subsets
Cardinality 3 = 10 subsets
Cardinality 4 = 5 subsets
Cardinality 5 = 1 subset

So, set S contains 32 subsets total, but I just want to find the number of subsets for M.
Thanks for your insight...I thought this would be easy, but can't crack it.
 A: These are the binomial coefficients: the number of $k$-element subsets of a set of size $n$ is $$n!\over k!(n-k)!,$$ or "$n\choose k$" (pronounced "$n$ choose $k$"). Here "$!$" is the factorial function.$^1$ You can derive this as follows:


*

*First, we count the sequences of $k$ distinct elements of our $n$-element set. The number of these is $n\cdot (n-1)\cdot ...\cdot (n-(k-1))$ - this is an ugly expression, but it's more snappily written as $$n!\over (n-k)!$$ (do you see why?).

*Now we need to "unorder" everything: the above is a terrible upper bound. Every set of $k$ elements can be listed in $k!$ many ways. So the actual number of sets is the number of sequences divided by the "overcounting" - that is, $${({n!\over (n-k)!})\over k!}={n!\over k!(n-k)!}.$$
As to the name, it's a good exercise to convince yourself that $n\choose k$ is the coefficient of the $x^k$ term in the expansion of $(x+1)^n$.
Incidentally, looking at the values of the binomial coefficients leads to a neat combinatorial picture - Pascal's triangle.

$^1$A convention which may seem odd at first, but is definitely the right definition: we set $0!=1$. So when $k=0$ we get $${n!\over k!(n-k)!}={n!\over 1\cdot (n-0)!}={n!\over n!}=1,$$  and similarly when $k=n$. If you think of the factorial function as being defined as "$a!$ is the number of distinct ways to order $a$ objects" (rather than "multiply all the whole numbers up to $a$") this makes sense - there's only one way to order no things. 
A: You effectively have to choose $M$ elements from a list of $N$ where order doesn't matter hence we have
$$\binom{N}{M}=\frac{N!}{M!(N-M)!}$$
possible choices. It is well known also that the set $S$ contains $2^N$ subsets in total.
A: Yes, by the binomial theorem the number of subsets with cardinality $k$ in a set with cardinality $M$ is $\binom{M}{k}$.
Let $\{x_k\}_{k=1}^n$ be the elements in the set $E$. Write any subset of $E$ asa a multiplication of elements from the subset, e.g. if $U=\{x_1,x_3,x_4\}\subset E$ then we will say $U\sim x_1x_3x_4$. Then the polynomial
$$(x_1+1)(x_2+1)(\cdots)(x_n+1)$$
has as its terms all possible subsets of $E$. Note that the total power of a term is the cardinality of its respective subset, e.g. $x_1x_3x_4$ has a total power of $3$ and the cardinality of $U$ is $3$. Finally, we determine the number of such subsets with cardinality $m$ by counting the number of terms in this polynomial with power $m$. Equating $x_i=x$ and applying the binomial theorem, we get
$$(x+1)^M=\sum_{j=0}^M\binom{M}{j}x^j$$
where $M$ is the cardinality of $E$. Thus, there are $\binom{M}{k}$ subsets of $E$ with cardinality $k$.
