Question

Can you explain the proof of the combination and why there is a "k-factorial" in the denominator.

Answer 1

This the formula I am talking about
\[n!/k! (n-k)!)\]

Answer 2

Consider the number of ways to order n objects in k places. In the first place, you could put any of the n objects. In the second place, you only have n-1 objects to put. If you continue, it is easy to see that there are n!/(n-k)! ways to do this.
Now, a combination is pretty much the same idea, except order doesn't matter. So what we want to know is, given a permutation of k objects, how many different ways can the order be switched around? If we know that, then we can divide out that number from each possible combination, or, using the distributive law, divide it out from our formula above.
So the number of ways to switch the order around is k!, and you can deduce that in a similar manner as I did above. Dividing that number out, you get n!/[k!(n-k)!]. QED

Answer 3

Let the number of combinations of n distinct things taken k at a time be x:

Now consider one of these x ways..
There are k ways in the selection which can be arranged in k! ways..

Thus, each of the x combinations will give rise to k! permutation..

So, x combinations will give rise to \(x \times (k!)\) permutations..

As a result, the number of permutation of n things taken k at a time will be \(x \times k!\) ..

But this is also equal to : \(^n P_k\)

So,

\[x \times k! = ^nP_k \implies x = \frac{^nP_k}{k!}\]

So,

\[x = \frac{n!}{(n-k)! \times k!}\]

Answer 4

Alright thanks so much guys