A standard pack of 52 cards consists of 4 suits, hearts, diamonds, clubs and spades. Each suit has 13 cards, from Ace to King. We deal randomly 5 cards from the deck of 52. 2 deals differing only by the order are considered the same.
How many different deals are there?
How many of them contain at most one ace?
How many contain cards of all four suits?
(52!)/(5!47!) or 52C5 = 2,598,960 different ways a 5-card hand can be dealt from a deck of 52 cards.
(4C1)x(48C4) = the amount containing at most one ace.
(5C4)x(47C1) = amount containing cards of all four suits. (not very sure of this answer)posted by Hmmm