Question: So, the problem I found goes like this:

> 
> You have $n$ different numbers, numbered from $ 1 $ to $n$. You can
>  randomly choose $m$ (different) of them. The computer also randomly
>  selects $m$ (different) of them. If you and the computer have exactly
>  $k$ common numbers, then you win a certain amount of money.
> 
> 
> 

The problem asks us to find the probability of winning.

I have solved some easier problems involving probabilities. But here, the only thing I could think of was that the probability for a certain sequence of $m$ numbers to emerge is:

$$ \frac{1}{\dbinom{n}{m}} $$

How do you solve it? I'm on my way of getting used to this type of problems and I could really use some help.

Answer: 
Good divided by total; or multiply your result with the number of matching sequences. There are $m\choose k$ ways to pick $k$ of the $m$ winning numbers and $n-m\choose m-k$ ways to pick the remaining numbers as non-winners. Divided by the total ways to pick $m$ numbres, we find
$$ \frac{{m\choose k}{n-m\choose m-k}}{n\choose m}$$
Let us assume you have picked your $m$ numbers. Now it's the computer's turn. It has to match $k$ of your numbers. Which $k$? These can be chosen in $\binom{m}{k}$ ways. Then it has to produce $m-k$ numbers which do not match any of yours. This can be done in $\binom{n-m}{m-k}$ ways.

So the number of ways the computer can match $k$ of your numbers is $\binom{m}{k}\binom{n-m}{m-k}$.

For the probability, divide $\binom{m}{k}\binom{n-m}{m-k}$ (the number of "favourables,") by the number of (equally likely) choices the computer can make. This is $\binom{n}{m}$.