# You are dealt five cards from an ordinary deck of 52 playing cards. In how many ways can you get a full house and a five-card combination containing two jacks and three aces?

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Oct 20, 2016

Full house = 3744, that specific full house = 24

#### Explanation:

To start, let's review what a standard deck of cards looks like: 13 ordinal cards (Ace, 2-10, Jack, Queen, King) - 1 of each ordinal in each of 4 suits (spades, clubs, hearts, diamonds), and so there are 52 cards: $13 \times 4 = 52$

There are ${C}_{52 , 5} = 2 , 598 , 560$ different possible hands with a 5 card poker hand. How many of those can be a full house? We'll break this one down by parts:

The first thing we need is 3 cards of the same ordinal, so we can express that as taking 1 of the 13 ordinals and getting 3 of 4 of them:

C_(13,1) xx C_(4,3)=(13!)/((12!)(1!))(4!)/((3!)(1!))=(13xxcancel12!)/((cancel12!)(1!))(4xxcancel3!)/((cancel3!)(1!))=13xx4=52

The next thing we need is 2 cards from the same ordinal and this ordinal has to be different than the one we just chose from, so that looks like:

C_(12,1) xx C_(4,2)=(12!)/((11!)(1!))(4!)/((2!)(2!))=(12xxcancel11!)/((cancel11!)(1!))(4xx3xxcancel2!)/((cancel2!)(2!))=12xx4xx3xx1/2=72

And now we multiply them together:

$52 \times 72 = 3744$

This is the number of full houses we can draw in a game of 5-card poker.

For the number of hands we can draw getting specifically 2 Jacks and 3 Aces, we calculate that this way - we only need to concern ourselves with picking out the number of cards of the 4 available in each of the listed ordinals, and so we get:

C_(4,3)xxC_(4,2)=(4!)/((3!)(1!))(4!)/((2!)(2!))=(4xx3!)/((3!)(1!))(4xx3xx2!)/((2!)(2!))=(4xxcancel3!)/((cancel3!)(1!))(4xx3xxcancel2!)/((cancel2!)(2!))=4xx4xx3xx1/2=24

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