Question #4b3a7

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!#.

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#