# Probability and counting without permutations and combinations.

I have been practising my probability skills and found a problem I wasn't able to solve. I would appreciate some help:

You have 20 horses. You randomly grab 5 of them, mark them and set them free. Later, you grab 4 of the total of horses and want to know what is the probability of the event ''exactly two of the horses you grabbed are marked'' ? The problem specifically says that you should solve it without using permutations and combinations.

I know it may be a silly problem, but I can't solve it. Thank you in advance for any help provided.

• What is the probability the first two you grab are marked and the others are not? What are the other possible patterns and their probabilities? What do you get when you get when you add up the probabilities? – Henry Oct 5 '19 at 6:55

Since the answer fundamentally involves combinations - it is $$\binom 52\binom {15}2\big/\binom{20}4$$ - it's not going to be possible to do it with a method which doesn't effectively come down to calculating combinations/permutations, even if you don't call them that.
You could explicitly calculate the probability that the first one you choose is marked, then the second one you choose (out of the remaining 19) is marked, then the next (out of the remaining 18) isn't, then the last isn't, but you still need to do this six times (in order to cover each possible combination of two marked horses out of the four you choose), so you're still really doing combinations even if you don't use them explicitly (and each of the six calculations involves the permutations $${}^{15}\mathrm P_2$$, etc, in disguise).