As an old bridgeplayer I choose to help you with b) 4!/(2!*1!*1!) * 13C5 * 13C5 * 13C3 * 13C0 / 52C13
Hope this gives you a clue to the other two /LWn "Peter Flom" <[EMAIL PROTECTED]> skrev i meddelandet news:[EMAIL PROTECTED] > A friend posed these three problems, which apparently relate to some > Chinese card game, but they are beyond what I remember of combinatorics > > If you deal 13 cards from a regular deck, what are the odds of getting > > a) 6 pairs > b) 5 cards of one suit, 5 of another, 3 of a third, and none of a > fouith? > c) Two series of 5 cards in straight order, and one series of 3 cards > in straight order? > > > Thanks in advance..... > > Peter > . > . > ================================================================= > Instructions for joining and leaving this list, remarks about the > problem of INAPPROPRIATE MESSAGES, and archives are available at: > . http://jse.stat.ncsu.edu/ . > ================================================================= . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
