document.write( "Question 137393This question is from textbook Fundmentals of Alegebraic Modeling
\n" ); document.write( ": If your class has 10 males and 15 females init, how many different committees consisting of 4 males and 3 females can be chosen? My answer is:
\n" ); document.write( "2.5 committees of males and 5 committees of females \n" ); document.write( "
Algebra.Com's Answer #100532 by edjones(8007)\"\" \"About 
You can put this solution on YOUR website!
We are dealing with combinations. We don't care about the order in which the people are chosen.
\n" ); document.write( "For the girls n=15, r=3
\n" ); document.write( "15C4=n!/((n-r)!*r!)
\n" ); document.write( "=15!/((15-3)!*3!
\n" ); document.write( "=(15*14*13*12!)/(12!*3*2)
\n" ); document.write( "=(15*14*13)/3*2
\n" ); document.write( "=5*7*13
\n" ); document.write( "=455 combinations of 4 girls.
\n" ); document.write( ".
\n" ); document.write( "For the boys n=10, r=4
\n" ); document.write( "10C4=10!/((10-4)!*4!)
\n" ); document.write( "=(10*9*8*7*6!)/(6!*4*3*2)
\n" ); document.write( "=(10*9*8*7)/(4*3*2)
\n" ); document.write( "=5*3*2*7
\n" ); document.write( "=210 combinations of boys
\n" ); document.write( ".
\n" ); document.write( "455*210
\n" ); document.write( "=95,550 different committees can be chosen.
\n" ); document.write( ".
\n" ); document.write( "Ed \n" ); document.write( "
\n" );