Question 128817This question is from textbook
: Find the number of possible 5-card hands that contain the cards specified: 4 aces and 1 other card. I was able to find the formula for a 5-card hand with 2 aces but I'm lost. For 2 aces, it's C(52,5) - C(48,5) - C(4,1) X C(48,4) but I don't understand all the parts. I know C(52,5) is the total possiblilities of a 5-card hand out of 52 cards and C(48,5) is the 48 remaining cards but I'm not sure about the C(4,1) and the next C(48,4)?
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the number of possible 5-card hands that contain the cards specified: 4 aces and 1 other card.
Pick 4 aces in 4C4 = 1 way
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Pick 1 other card in 48C1 = 48 ways
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Total # of hands = 1*48 = 48
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I was able to find the formula for a 5-card hand with 2 aces
Pick 2 aces in 4C2 = 6 ways
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Pick 3 other cards from 50 in 50C3 = 19600 ways
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Total # of hands = 6*19600 = 117600 ways
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Cheers,
Stan H.
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