I'm trying to figure out how many groups of 8 lottery numbers can be formed from the set $A=\{1,2,3,\dotsc,45\}$. The first 3 numbers are "winning" numbers and the left over 5 "losing" numbers.
I understand that I need to figure out how many groups of 8 can be formed from 45 elements.
$$\binom{45}8 = \frac{45!}{8!(45-8)!}.$$
What I'm having a problem with is understanding how to take these rules into account.
Maybe something like:
$$\binom{45}3 * \binom{42}5?$$
To be honest I'm quite lost. All hints and pointers are appreciated.