Permutations and Combinations maths C problem.?

Question

In poker you are dealt 5 cards. The dead mans hand consists of 2 pairs - Aces and eights. In how many ways can you be dealt this hand?

Using a regular pack of 52 cards.

Answer 1

Hi,



To get a "Dead Man's Hand" with 2 Aces and 2 eights, there are 6 ways to get 2 Aces, 6 ways to get 2 eights, and 44 other choices for the 5th card.

6 x 6 x 44 = 1,584 ways to get 2 Aces and 2 eights with any other different 5th card



In Wild Bill Hickok's famous "Dead Man's Hand", the 5th card was rumored to have been the 5 of diamonds. This exact hand could come from 6 ways to get 2 Aces, 6 ways to get 2 eights, and just one choice for the 5th card.

6 x 6 x 1 = 36 ways to get a hand like Wild Bill's.



If you simply want to get 2 pairs of any 2 cards with a different 5th card, then there are 13 different cards that could be chosen for pairs. So there are 13nCr2 = 78 different combinations of 2 different cards, like queen-three or ten-jack.



Once the type of cards chosen was selected, there are 6 different ways to select 2 of those 4 cards from 4nCr2 for both types.



The fifth card could be any of the remaining 44 cards in the deck that are neither of the previously chosen type of cards.



So any hand with 2 pairs could be made in:

13nCr2 * 4nCr2 *4nCr2 * 44 = 78 * 6 * 6 * 44 = 123,552 hands with any 2 pairs



I hope that helps!! :-)

Answer 2

We, count all ways we can get the cards, minus the combination where we don't get pairs of Aces and eights, there are 3 cases:

1. Only pair of Aces

2. Only pair of Eights

3. No pair of Aces or eights.



The total combination we could get is 52C5



For the first case, we have 3 cases:



case 1. There are two Aces

in this case, we also have two cases:



a. there is exactly one Eight

then the number of combination is 4C2(for the Aces) * 4(the Eight) * 44 * 43 (for the other two cards, as Aces and Eights can't be used anymore)



b. There is no eight.

then the number of combination is 4C2(for the Aces) * 44 * 43 * 42( for three other cards, no aces or eights)



The total number on case 1 is 4C2 * 44 * 43 * 46



case 2: There are three Aces

Similar with previous case, we have two cases:



a. There is exactly one Eight

then the number of combination of this case is 4C3(for the Aces) * 4(for the eight) * 44(for the last card, no Aces or Eights)



b. There is no Eights

then the number of combination of this case is 4C3(for the Aces) * 44 * 43 (for the other two cards, no Aces or Eights)



So total number of combination in case 2 is 4C3 * 44 * 47



case 3: There are 4 Aces

we have the number of combination is 4C4(for the Aces) * 48(for the last cards, no Aces)



Then total number of combination on case "There are only pair of Aces" is 6 * 44 * 43 * 46 + 4 * 44 * 47 + 1 * 48, call it as X



For the case "there are only pair of eights", we have the same calculation with previous case, so the total is also X.



For the case "There are no Aces or Eights" we have the number of combination is 44 * 43 * 42 * 41 * 40 (5 cards with no Aces or Eights)



So the answer to your question is 52C5 - 2X - (44 * 43 * 42 * 41 * 40)

Answer 3

If a word as one letter, there is one permutation, that's one if there are 2 letters there are 2 variations, or one situations 2 although, you presently have 2 one letter variations, that's the single situations 2 if there are 3 letters, you have have been given six variations, or one events 2 situations 3 yet you presently have 3 one letter adjustments, or one events 3 now it may become clean you probable have 4 letters you are able to have 24 adjustments, or a million events 2 situations 3 events 4 with 4 one letter variations, or one time 4 and eight 2 letter adjustments, or one events 2 situations 4 so, casserole has 9 letters which provides you with 9*a million*2*3*4*5*6*7*8 plus 9*a million*2*3*4*5*6*7 plus 9*a million*2*3*4*5*6 etc, except you get to 9*a million the equation may well be 9! + 9(a million!+2!+3!+4!+5!+6!+7!) 8! isn't blanketed in the parenthesis once you think approximately that that's an element of 9!

Answer 4

//

4C2*4C2*44=1584