document.write( "Question 880947: 5P3 \n" ); document.write( "
Algebra.Com's Answer #532260 by Edwin McCravy(20055)\"\" \"About 
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document.write( "5P3 = 5*4*3 = 60\r\n" );
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document.write( "Start with the first number, and start multiplying by 1 less each time\r\n" );
document.write( "until you have the second number of factors.  Then multiply.\r\n" );
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document.write( "We started with 5, then multiplied by 1 less, or 4, then by 1 less than\r\n" );
document.write( "that, 3. So we had 5*4*3. We stopped there because we had 3 factors.\r\n" );
document.write( "Then we multiplied them and got 60.  \r\n" );
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document.write( "Edwin
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