document.write( "Question 429255: In how many ways can three aces be drawn from a standard deck of 52 cards?\r
\n" ); document.write( "\n" ); document.write( "MY WORK SO FAR:
\n" ); document.write( "Im not sure which approach to take on this problem.
\n" ); document.write( "52 cards=n
\n" ); document.write( "3 aces=r\r
\n" ); document.write( "\n" ); document.write( "52!/(52-3)!
\n" ); document.write( "52!/49!
\n" ); document.write( "52x51x50x49x.../49x48x47x...
\n" ); document.write( "52x51x50=132,600\r
\n" ); document.write( "\n" ); document.write( "OR\r
\n" ); document.write( "\n" ); document.write( "I know there are 52 cards in a deck, 4 suits in a deck with one ace each, so 4 aces in a deck.
\n" ); document.write( "so...
\n" ); document.write( "n=4
\n" ); document.write( "r=3
\n" ); document.write( "4!/(4-3)!
\n" ); document.write( "4!/1!
\n" ); document.write( "4x3x2x1=24\r
\n" ); document.write( "\n" ); document.write( "I need help! \n" ); document.write( "

Algebra.Com's Answer #298279 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
Your first \"solution\" actually suggested the number of ways to pick three of *any* card out of a deck of 52 cards, where the order does matter.\r
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\n" ); document.write( "\n" ); document.write( "The second solution is closer, you picked three aces out of four cards, using the expression 4P3. However this over-counts quite a bit, because 4P3 assumes the order also matters. Since we are choosing three aces and the order is assumed not to matter, it would be represented as 4C3, or 4!/(3!1!), which is equal to 4. \n" ); document.write( "

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