When considering the suit of the cards, 5 from 52 cards have 2598960 different combinations.
But how many different combinations are possible when neither suit nor order of the cards are considered? (E.g. 2d, 3s, 4h equals 4s, 3c, 2c)
2 Answers
quads: 13 * 12 = 156
full houses: 13 * 12 = 156
trips: 13 * 12 * 11 / 2 = 858
two pairs: 13 * 12 * 11 / 2 = 858
single pair: 13 * 12 * 11 * 10 / 6 = 2860
5 different cards: 13 * 12 * 11 * 10 * 9 / 120 = 1287
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total: 6175
exemplary explanation for single pair:
13 possibilities for the rank of the pair, 12 possibilities for the first single card (the rank of the pair is not allowed), 11 possibilities for the rank for the second single card (ranks of pair and first single card are not allowed), 10 possibilities for the rank for the third single card. As the order of the cards is not considered, the resulting number 13*12*11*10 needs to be divided by the number 3! = 6 of permutations of the 3 single cards.
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This looks correct, however could you please explain why with trips, two pairs and single pair you divide by 2, 2 and 6 respectively?– RaymondJan 19, 2019 at 17:14
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1@RaymondTimmermans thanks for your feedback. I've added an explanation for the single pair to my answer. For two pairs we divide by two to make up for swapping the rank of the pairs. For trips we divide by two to make up for swapping the rank of the two single cards.– azimutJan 19, 2019 at 18:51
My answer is incorrect. It addresses the amount of permutations, not combinations.
Ignoring suits makes things actually way easier, not harder. When suits don't matter you basically have thirteen possibilities on each card, five times. Equaling 13^5 = 371293 permutations. However there are a few permutations in here that are not possible and that is five of a kind. So we substract all five of a kind permutations, which are 13 in total, to get to our answer. 371280.
