Probability
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HallsofIvy
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That looks fairly straight forward. First put the four players in a specific order. There are 4! ways to do that. Now give the Ace, King, Queen, and Knave of one suit to each person. There are 4! ways to distribute the four suits among the four people. That leave 36 cards to be distributed at random, 9 cards to each person. There are \(\displaystyle _{36}C_9= \frac{36!}{9! 27!}\) ways to sort out one hand of 9 cards. Then there are \(\displaystyle _{27}C_9= \frac{27!}{9!18!}\) ways to sort out the second hand and \(\displaystyle _{18}\frac{18}{9!9!}\) ways to sort out the third 9 card hand, leaving the fourth hand.
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Yes, except there are suits and each player is dealt a four-card flush (Jack, Queen, King, Ace).
So, if your deck is four colors (1 through 13 in each color), then the question is: how many different deals are possible where each player ends up with 1 through 4 of the same color?
Checked my coding...99.9% sure correct...
Send me your revised code, if you like. I'll recode in justBASIC, justFERFUN.
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It was given in one of the EAMCET material. But i dont know the correctness in the answer.