Mark M. answered • 05/29/15
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For a combination (order does not make a difference) of three kings
4! / (3!(4-3)!) = 4
For a combination of the remaining 49 cards
49! / 2!(49-2)!) =
(49 · 48) / 2 =
1176
The total possibilites are
4 · 1176
4704
Mark M.
The condition was exactly three kings. The fourth king cannot be used.
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05/29/15
Michael W.
Your formula says to choose 3 out of the 4 kings, and then two random cards from the other 49. One of those other 49 cards is the remaining king, so the formula you gave would allow the fourth king to accidentally be picked. That's exactly the issue.
To avoid choosing the last king when we pick the last two cards in the hand, we can only choose from 48 cards in the deck...the non-kings. That's 48-choose-2, not 49-choose-2.
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05/29/15
Mark M.
Understood, thank you.
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05/29/15
Michael W.
05/29/15