2006-09-17 13:51:23 · 8 answers · asked by tatumt501 1 in Science & Mathematics ➔ Mathematics
311,875,200 possible hands...
52*51*50*49*48
2006-09-17 14:06:07 · answer #1 · answered by MB_Bailey 3 · 1⤊ 0⤋
First you have to consider that in a "card hand" the order does not matter.
This reduce the problem to evaluate the number of subsets of a set with 52 elements.
Subsets of a set are Combinations.
To evaluate the number of combinations of k elements from a set with n elements we use the formula: C(n,k) = n!/k!(n-k)!
So, the answer is: C(52,5) = 52!/5!(52-5)! = 52!/(5!47!)=
52x51x50x49x48/(5x4x3x2) = 52x51x10x49x2 = 2,598,960
2006-09-17 14:20:19 · answer #2 · answered by vahucel 6 · 1⤊ 0⤋
A simple formula for this is:
"n-Choose-r"
In your case, 52-choose-5
Here's the formula:
n-Choose-r = (n-factorial) / [ (r-factorial) * ( (n - r) factorial) ]
stated another way:
n-C-r = (n!) / (r! * (n - r)! )
n-factorial means n*(n-1)*(n-2)*(n-3)*...*3*2*1
For instance, 5! = 5*4*3*2*1
So, 52-C-5 = (52!) / [(5!) * (47!)]
Expand to simplify:
(52*51*50*49*48*47*46*45*...)
/ [(47*46*45*...) * (5*4*3*2*1)
=(52*51*50*49*48) / (5*4*3*2*1)
=(13*10*17*49*24)
=2,598,960 possible hands.
2006-09-17 16:47:14 · answer #3 · answered by Anonymous · 1⤊ 0⤋
10
2006-09-17 13:52:42 · answer #4 · answered by uceuce 2 · 0⤊ 2⤋
There are 2,598,960 possible 5-card hand in a 52-card deck.
2006-09-17 13:57:42 · answer #5 · answered by Michael 4 · 1⤊ 0⤋
Wouldn't it be 52x51x50x49x48? 311,875,200 keeping suits in consideration - 52 possible cards times 51 left etc.
2006-09-17 14:00:29 · answer #6 · answered by DrJunk 3 · 1⤊ 0⤋
As many as you want
2006-09-17 13:53:45 · answer #7 · answered by Anonymous · 0⤊ 1⤋
10.4
2006-09-17 13:53:39 · answer #8 · answered by elmo76570 1 · 0⤊ 2⤋