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Math help please....?

A poker hand consists of 5 cards dealt from an ordinary deck of 72 playing cards. (a). How many poker hands are possible? (b). How many different hands consisting of three kings and two queens are possible? (c). The hand in part(b) is an example of a full house: 3 cards of one denomination and 2 of another. How many different houses are possible? (d). Calculate the probability of being dealt a full house? oops its 52 cards!!

Public Comments

1. go here www.algebra.com they are whizzes and can explain to you in detail of how they received the answer.
2. only 52 cards in a deck
3. well, an ordinary deck of playing cards is 52 cards, not 72.
4. this is a probability problem, so the poker hands would be 72 x 71 x 70 x 69 x 68 because it is how many cards are possible... does that help? and same for b and i guess c and well d too, email me if you need more then that...
5. I'll answer part A... 13991544( and im being serious..)
6. don't ask us, ask bunch of geeks who actually do there hw
7. Please! You have to ask a correct question, to get a correct answer! There are only 52 cards in a complete deck... Or is this a special deck??
8. the answer is Secret!! lol. do your own homework pls.
9. 52 cards actually.
10. Candy, why do I feel as though I am doing your homework for you? Anyhow, let's make the deck 52 cards instead of 72. That's a bit more 'regular'. The number of ways to take 5 things from 52 without the order being important (in other words, ace of spades delt 1st or 3rd doesn't matter if it ends up in your hand either way) is: n!/ ((n-s)! x s!) or, for this problem 52!/((52-5)! x 5!) or, 52!/(47! x 5!). To cut to the chase, it works out to be 52x51x50x49x48/120=2,598,960 different hands.
11. good luck with that.
12. I hope you mean 52 cards. Anyway, to figure out this problem, you would need to find the number of combinations of 5 taken from 52. The equation for that is: 52!/(52-5)!5! This is equal to 52!/47!5! Which is equal to: (48x49x50x51x52)/(2x3x4x5) There are a total of 2,598,960 poker hands. It's best to find the number of ways to get three kings, then two queens, then to multiply the two. There are four possible kings and combinations of three. 4!/(4-3)!3! This equals 4. 4!/(4-2)!2! This equals 6. 6x4=24 There are 24 possibilities of three kings and two queens. (c) is a little harder to explain. For every card Ace thru King there are 12 other cards it could make a full house with. The king can make 24 with a queen, but it can also make 24 with ace, two, three, four etc. To find the total number one must multiply 24 by 12. This equals 288. Now we must multiply that number by the total number of different kinds of cards in the deck. That would be 13. 288x13 = 3,744. There are 3,744 ways to make a full house. To find the probability, one must use the answer from question (a) and put the answer to question (c) over it. 3744/2598960 = 6/4165. Or an approximately 0.144 % chance of being dealt a full house.
13. Check out this question again. http://answers.yahoo.com/question/?qid=1006050124394&r=w#NbUvWjS7VTU1SSpDbLZf My answer was wrong before, but now I have the correct answer.
14. a 52C5= 52!/(5!47!)=2 598 960. b There are 4 ways you can select three kings and 6 ways you can select two queens, so there are 6x4=24different 3 kings+2queens hands. c There are 13 denominations.For each denomination there are 4 ways you can have 3 cards and 6 ways you can have 2 cards, so there are4x6x13=312 ways you can have a full house. d There are 2 598 960 different poker hands, of which 312 are full houses, so your chance of being dealt a full house is 312/2 598 960 = 1 in 8330.
15. a.) 2,598,960 b.) 24. (4 possibilities of 3 Kings times 6 possibilities of 2 Queens) c.) 3,744 (4 possibilities of 3 of a kind suit combinations times 6 possibilities of 2 of a kind suit combinations times 13 possibilities of 3 of a kind values times 12 possibilities of 2 of a kind values) d.)1 in 693