In Lotto 6 different Numbers from the Set M = {1, 2 ..., 45} are to be guessed. How many guesses with exactly k correct numbers are present?

Question

So I have a solution for this exercise:

$\binom{6}{k}$ * $\binom{39}{6-k}$

I just have troubles understanding why it works. So the left part is where you "choose" how many numbers you want to be correct which you then multiply (why?) by the amount of different combinations from the remaining 39 numbers if you want to choose unique combinations of 6 minus the amount of numbers you want to be correct.

Hope someone can help me

Kind regards

Answer 1

There are a total of $45$ elements out of which only $6$ are "correct" elements.

So, there are $45-6=39$ "incorrect" elements.

In each guess, you have to select $6$ elements. You want $k$ "correct" elements, so there are $6-k$ "incorrect" elements.

So, you choose $k$ elements from $6$ elements and $6-k$ elements from $39$ elements and then apply the rule of product.