Question

Question #4b3a7

Answer 1

`4512` ways if the order in which the cards are drawn does not matter.

`541440` ways if the order in which the cards are drawn does matter.

Explanation:

For this problem, we will use the following:

• The multiplication principle, which states that if there are `m` ways of doing one thing and `n` ways of doing another, there are `mn` ways of doing both.

• Choose notation, `((n),(k)) = (n!)/(k!(n-k)!)`, which calculates the number of ways of choosing a subset of size `k` from a set of size `n`.

Depending on if we also are distinguishing between drawing the same cards in a different order, we may also use that the number of permutations of a set of `n` objects is given by `n!`.

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Proceeding, note that there are `((4),(3))` ways of selecting three of the four aces to be in the hand, and then `((48),(2))` ways of selecting two more cards from the forty eight remaining non-ace cards in the deck. Thus, by the multiplication principle, the total number of hands is given by

`((4),(3))*((48),(2)) = (4!)/(3!1!) ((48!)/(2!46!))`

`=4*(48*47)/2`

`=4512`

If the order in which the cards are drawn does not matter, then we are done. If it does, then we note that for each of the above hands, there are `5!` different orders (permutations) in which the cards can be drawn, meaning the total becomes

`4512*5! = 541440`