document.write( "Question 129659: Laura is training for a week-long mountain cycling tour. She has 12 short hilly routes from which to choose mid-week rides.
\n" ); document.write( " a) How many ways can she choose 4 different rides from the list for the first week's training if order matters?
\n" ); document.write( " b) How many ways can she choose 4 different rides if order does not matter?
\n" ); document.write( " c) If she has chosen the first weeks rides, how many ways can she choose four more different rides for the second week? Assume that order does not matter
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Algebra.Com's Answer #94673 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Laura is training for a week-long mountain cycling tour. She has 12 short hilly routes from which to choose mid-week rides.
\n" ); document.write( "a) How many ways can she choose 4 different rides from the list for the first week's training if order matters?
\n" ); document.write( "Ans: 12!/(12-4)! = 12*11*10*9 = 11800
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\n" ); document.write( "\n" ); document.write( "b) How many ways can she choose 4 different rides if order does not matter?
\n" ); document.write( "Ans: 12C4 = [12*11*10*9]/[1*2*3*4] = 495
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\n" ); document.write( "\n" ); document.write( "c) If she has chosen the first weeks rides, how many ways can she choose four more different rides for the second week? Assume that order does not matter
\n" ); document.write( "Ans: 11C4 = 330
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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