document.write( "Question 1000939: How many different ways can a club choose a committee of 4 from 15 members? \n" ); document.write( "
Algebra.Com's Answer #618254 by Alan3354(69443)\"\" \"About 
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How many different ways can a club choose a committee of 4 from 15 members?
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\n" ); document.write( "The 1st choice is 1 of 15.
\n" ); document.write( "Then 14, 13 & 12
\n" ); document.write( "--> 15*14*13*12 = 32760 ways
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\n" ); document.write( "If there is no difference between the members, then A,B,C&D is the same as C,B,A,&D, etc.
\n" ); document.write( "--> 4*3*2*1 = 24 different ways the can be chosen.
\n" ); document.write( "32760/24 = 1365 different ways.
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