SOLUTION: In the card game Crazy Eights, how many different eight-card hands can be dealt from a standard 52- card deck?

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Question 1196819: In the card game Crazy Eights, how many different eight-card hands can be dealt from a standard 52- card deck?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13198) (Show Source):
You can put this solution on YOUR website!

From the basic definition of "n choose r", the answer is "52 choose 8":

Use a calculator to find the answer is 752538150.

Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!

n = 52 cards
r = 8 selections
The order doesn't matter

Use the nCr combination formula

n C r = (n!)/(r!(n-r)!)
52 C 8 = (52!)/(8!*(52-8)!)
52 C 8 = (52!)/(8!*44!)
52 C 8 = (52*51*50*49*48*47*46*45*44!)/(8!*44!)
52 C 8 = (52*51*50*49*48*47*46*45)/(8!)
52 C 8 = (52*51*50*49*48*47*46*45)/(8*7*6*5*4*3*2*1)
52 C 8 = (30,342,338,208,000)/(40,320)
52 C 8 = 752,538,150

Answer is 752,538,150
(approximately 752.5 million)