52 poker cards are evenly distributed to four people, how many ways can be made if two of the people get the same number of Ace? (2,2,0,0) & (1,1,2,0)
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Originally Posted by ddmmkk 52 poker cards are evenly distributed to four people, how many ways can be made if two of the people get the same number of Ace? (2,2,0,0) & (1,1,2,0) Is zero a number? Are the four people distinguishable? CB
(2,2,0,0) & (1,1,2,0) mean there are two chances of two people gettig the same number of Ace. For example, A gets 2 and B gets two. Others have none.
Originally Posted by ddmmkk (2,2,0,0) & (1,1,2,0) mean there are two chances of two people gettig the same number of Ace. For example, A gets 2 and B gets two. Others have none. Number of As for players 1,2,3,4 such that 2 or more have the same number: (0,0,0,4) (0,0,1,3) (0,0,3,1) (0,0,4,0) (0,4,0,0) (4,0,0,0) (0,1,0,3) (0,3,0,1) : : ?
o i see it! but how about if the number of ways of two people getting 2 Ace , and others get none?
Originally Posted by ddmmkk o i see it! but how about if the number of ways of two people getting 2 Ace , and others get none? (2,2,0,0) (2,0,2,0) (2,0,0,2) (0,2,2,0) (0,2,0,2) (0,0,2,2) and ask yourself if which As are where count as separate ways. CB
Originally Posted by ddmmkk 52 poker cards are evenly distributed to four people, how many ways can be made if two of the people get the same number of Ace? Question: Are there any ways for at least two of these four people would not get the same number of Aces?
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