A suit contains 13 cards of the same kind. 4 cards may be choosen out of 13 in 13C4 (715) ways. There are 4 suits. Therefore, the number of possible hands for getting 4 cards of the same suit is 4 x 13C4 = 4 x 715 = 2,860.
2,560
If the cards are all different then there are 13C7 = 1716 different hands.
4(13c4)(39c1) = (4)(715)(39)=1,11,540
13 x 12 x 11 x 49 x 48 13 x 12 x 11 because there are 13 possible cards for any given suit, then 12 more of the same suit, then 11 more for the same suit. At this point, you have 49 cards left, then 48. So there are 4,036,032 possible hands like that.
23
The number of different 3-card hands that can be dealt from a deck of 52 cards is given by the combination formula, which is: C(52, 3) = 52! / (3! * (52 - 3)!) = 22,100 Therefore, there are 22,100 different 3-card hands that can be dealt from a deck of 52 cards.
The odds of getting 4 cards in any particular suit (for example, hearts) would be:1/4*1/4*1/4*1/4=1/256.The odds of getting 4 cards of the same suit (any suit) would be 4x greater:1/256*4=1/64Because it doesn't matter what the last card is, the probability is 1/64.
If the hand must contain three 8's and to cards that are not 8's - the total number of possibilities is 2801.
If you mean all the cards have to be face cards, then it would be a four of a kind, full house, three of a kind, and two pair. For example: Four of a kind, four Jacks and a Queen. Full house, Kings over Queens. Three of a kind, three Jacks, a Queen, and a King. Two pair, a pair of Kings, a pair of Jacks, and a Queen. If you mean how many different combinations can be made using only only face cards, there are 95040 possible hands.
three
This is a permutation, which is from 52 select 3, or52P3, or 132600. The other way to think of this is you have 52 choices for the first card, 51 choices for the second card, and 50 choices for the third card. Therefore you have 52*51*50 possible hands or 132600.
26 of the 52 cards are black. For each of those cards, the second card in your hand may not be the same card, but it may be any of the 25 remaining ones. The third card, likewise, can not be the first two, but can be any of the 24 remaining. For a four card hand that's 26 * 25 * 24 * 23 = 358800 possible hands However, that number includes sets that are effectively duplicates (1,2,3,4 and 4,3,2,1 are both accounted for in that number, but for our purposes those are the same hand). We can arrange each set of cards 4 * 3 * 2 * 1 = 24 ways. So to remove those possibilities we take 358800/24 = 14950 possible hands of all black cards.