Cards, a deck of 36 cards 1 ace and one of hearts?

Question

Combinatorics

Hello)
From a deck of 36 cards, 5 cards were randomly taken. In how many ways can there be
1 ace and 1 card of the suit of hearts among these 5 cards?
Preferably with an explanation

Answer 1

If your task is that the cards chosen at random contain at least one ace and at least one heart, then you need to find Count(>0 hearts AND >0 aces). You can reverse this by counting all bad choices by subtracting them from all choices:
Count(>0 of hearts AND >0 of aces) = Count() - Count(0 of hearts OR 0 aces)
Further, COUNT(A or B) can be decomposed into Count (A) + Count(B) - count(A AND B).
The final formula for the answer is:
Count() - Count(0 of hearts) - Count(0 of aces) + Count(0 of hearts AND 0 aces)
In fact, this is an inclusion-exclusion formula. But in the final formula, everything is easy to count:
Count() = C(5,36) - all options: combinations of 5 out of 36.
Count(0 aces) = C(5, 32) - you can’t take aces
Count(0 of hearts) = С(5, 27) - you cannot take 9 worms
Count(0 hearts AND 0 aces) = C(5, 24) - you cannot take 9 hearts and 3 remaining aces.
Count through fatorials and add with the correct signs.
If the task is exactly one ace and exactly one heart, then there are 2 options. Either the ace of hearts is taken, or they are two different cards.
In the first case, the remaining 4 cards are any of the 24 non-aces, non-hearts, i.e. this part is C(4,24). In the second case, you take one of the 3 aces, one of the 8 of hearts, and the remaining 3 of the non-aces, non-hearts, i.e. the answer is 3*8*C(3.24). Sum both parts.