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A friend of mine and I have our own brand of Cribbage where we throw extra Cards into the game hypothetically speaking how many 15s would there be if there was five fives in one hand

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Three fives create a single fifteen.

Four fives create four fifteens: you have four choices of which card not to use.

Five fives create ten fifteens: you have five choices of the first card and four choices of the second card, but half of those are equivalent to the other half, so there are ten pairs of cards to not use.

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    Also, (5 choose 3) = 10. – The Chaz 2.0 Feb 13 '18 at 18:37
  • That requires dealing with 3! to effectively explain in the same way as above, this way lets me use the half-half. They're equivalent by symmetry and by double-counting anyway. @TheChaz2.0 – Nij Feb 13 '18 at 18:59
  • So if five 5s makes ten 15s how many pairs would it make. – Jake Kincade Feb 13 '18 at 19:53
  • It would also make ten pairs for the same argument as the choice of excluded cards. In each case you're choosing two cards out of five and ignoring the order, the only difference is which set you consider for scoring. Hence the statement about symmetry and double-counting. – Nij Feb 13 '18 at 21:47
  • (regarding the first comment) - I don't buy the whole "avoiding factorials" argument. It just so happens that we can consider an equivalent calculation that uses 2! = 2, but for a direct approach (or a hand with six cards), we would invoke the binomial coefficient. – The Chaz 2.0 Feb 14 '18 at 2:31

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