One jack and one club

Question

How many ways a player can be dealt exactly one jack and exactly one club ? Of course, A player can be dealt 5 cards in a normal poker game

So we have 4 different jacks and 13 different club cards. However, one of the jack is already a club.

The way I think is that I just choose one of the 4 jack cards and then 1 of the 12 club cards (I excluded here the jack). And now we have $$52 - (4 + 12) = 36$$ different cards to choose 3 cards from.

so the answer should be $${4 \choose 1} \times {12 \choose 1} \times {36 \choose 3}$$

But then I may have chosen a jack of clubs, then we are not allowed to have another club. How can I account for this ?

Answer 1

Either:

• You get the jack of clubs, and 4 cards which are neither jacks nor clubs, or:
• You get a non-club jack, a non-jack club, and 3 cards which are neither jacks nor clubs.

These are disjoint possibilities. Count each of them up separately.