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Combinatorics Problem |
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| Apr10-05, 02:05 PM | #1 |
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Combinatorics Problem
Suppose you play a game of cards in which only four cards are dealt from a standard deck of 52 cards. How many ways are there to obtain three of a kind? (3 cards of the same rank and 1 card of a different rank, for example 3 tens and 1 queen.)
Could someone help me with how to do this problem? I tried doing 4C3 x 13 x 52C1, which was obviously wrong. :/ Any help would be appreciated. |
| Apr10-05, 02:24 PM | #2 |
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You can choose first card in 52 different ways. You can choose the next two (same of a kind as the first one) in 3 * 2 ways. And the last card, in 49 different ways.
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| Apr10-05, 10:34 PM | #3 |
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Recognitions:
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For any given rank, there are C(4,3) combinations of 3 cards chosen from the rank's 4 cards. Since there are 13 different ranks, a total of {13*C(4,3)} possible combinations of {3 cards from the same rank} exist. Finally, there remain {(52 - 4) = 48} cards in the 12 other (different) ranks from which to choose the final card. Hence: {Total Combinations of "3-of-a-Kind" from std 52 Cards} = {13*C(4,3)}*(48) = (2496) ~~ |
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