How do you do this problem?
You are dealt five cards from an ordinary deck of 52 playing cards. In how many ways can you get
a) a full house and b) a five card combination containing two jacks and three aces (a full house consists of three of one kind and two of another. For example, A-A-A-5-5 and K-K-K-10-10 are full houses.)
How do you do this problem?
Comments
Think about it like an organization problem. Assume that I have enough cards to make every possible full house, and I just want to come up with a good way of organizing them. So first we sort them by what value shows up 3 times. I would need 13 categories to do this sort, I can lay them out in order Ace to King (or 1 to 13 if you like).
Next, for the sake of my analogy, I will put a sorting on the suites, the order I choose is Spades, Clubs, Hearts, Diamonds, but which order you choose doesn't matter. Now take piles you made before and sort them by which suit is missing, so you get 4 sub-piles within each pile.
You can see that at this point you've sorted all the possible Full House draws into 13*4 piles. Next we sort each of these piles by the 2 card suite. First we need 12 sub-piles representing the 12 values which are left (since you can't have 3 aces and 2 aces). Next we need to order by which suites are missing, or equivalently by which ones are present. Combinatorics tells us that this is 4 choose 2 where order doesn't matter, or in other words 6. At this point each pile contains only 1 full house, so if I count the number of piles I've counted the total number of full houses.
So there are 13*4*12*6 = 3744 hands yielding a full house.
Now part b is easier, they've eliminated the need for piles based on the value of the card. There are only 4 ways to draw 3 aces from a deck, since there are only 4 ways to leave 1 ace in a deck. There are 6 ways to draw 2 Jacks. So there are 6*4 or 24 hands yielding the combination in part b.
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