document.write( "Question 128817This question is from textbook
\n" ); document.write( ": 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)?
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Algebra.Com's Answer #94217 by stanbon(75887)\"\" \"About 
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Find the number of possible 5-card hands that contain the cards specified: 4 aces and 1 other card.
\n" ); document.write( "Pick 4 aces in 4C4 = 1 way
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\n" ); document.write( "Pick 1 other card in 48C1 = 48 ways
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\n" ); document.write( "Total # of hands = 1*48 = 48
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\n" ); document.write( "\n" ); document.write( "I was able to find the formula for a 5-card hand with 2 aces
\n" ); document.write( "Pick 2 aces in 4C2 = 6 ways
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\n" ); document.write( "Pick 3 other cards from 50 in 50C3 = 19600 ways
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\n" ); document.write( "Total # of hands = 6*19600 = 117600 ways
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\n" ); document.write( "\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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