Need help on this one,
Let A be a finite set with cardinality |A| = n. Prove that A has exactly
C(n,k) subsets ofcardinality k. ( n and k are arbitrary integers such that
0 =< k = < n)
Hi baz,
If we take k elements from n in all possible arrangements of the k elements,
there will be
$\displaystyle n(n-1)(n-2)(n-3)........(n-[k-1])$ many arrangements.
as there are k factors in the above expression,
since by arranging them,
there are n choices for the 1st element,
having chosen one there are (n-1) remaining choices for the 2nd,
(n-2) remaining choices for the 3rd....etc
Since, we are not multiplying by $\displaystyle (n-k)((n-[k+1])(n-[k+2]).....3(2)1$
we can write that as
$\displaystyle \frac{n!}{(n-k)!}$
This has counted all possible groups of k distinct elements from the n
in all possible orders
hence we need to unarrange them, to find the number of subsets
with k elements.
Therefore we divide by k!
$\displaystyle \frac{n!}{(n-k)!k!}$ is written $\displaystyle \binom{n}{k}$ or $\displaystyle C(n,k)$ or $\displaystyle Nc_k$
The number of ways to choose k elements from a total of n is $\displaystyle \binom{n}{k}$