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June 16th, 2012, 05:07 AM   #31
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Re: Card fun

Quote:
 Originally Posted by aswoods The generating function solution is to find the coefficient of x^5 in $(1+y(3x+3x^2+x^3))^5(1+x)^6$: $243y^5+4050y^4+9990y^3+5460y^2+600y+6$ which gives 243+4050+9990 = 14283 and 5460+600+6 = 6066.
And how do you know how to group the numbers into those summing to winners?

 June 16th, 2012, 05:59 AM #32 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Card fun For each of the five kinds of low card, you can either pick none (x^0 = 1), or one card in three ways (3x^1), two cards in three ways (3x^2), or three cards in one way (1x^3). The "y" is used as a tag to indicate that one kind of low card has appeared. For each of the six kinds of high card, you can either pick none (x^0) or one (1x^1). Now you are interested in hands of 5 cards, so you expand and look at the coefficient of x^5. The degree of y indicates how many different kinds of low card have been picked. There are 243 five-card hands in which 5 kinds of low card appear, 4050 five-card hands in which 4 kinds of low card appear, etc.

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