document.write( "Question 951901: 33 people distributed among 3 committee of 11. How may ways can this be done? \n" ); document.write( "
Algebra.Com's Answer #581723 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
\r\n" );
document.write( "First we'll calculate as though the committees are ordered: committee A,\r\n" );
document.write( "committee B, and committee C.  Then we'll \"unorder\" the committees by dividing\r\n" );
document.write( "by 3!, the number of orderings of the 3 committees.\r\n" );
document.write( "\r\n" );
document.write( "We can choose the 11 people for committee A in 33C11 = 193536720 ways.\r\n" );
document.write( "We can then choose the 11 people for committee B in 22C11 = 705432 ways.\r\n" );
document.write( "Then the 11 remaining people make up committee C in 11C11 = 1 way.\r\n" );
document.write( "\r\n" );
document.write( "That's (33C11)(22C11)(11C11) = 136526995463040\r\n" );
document.write( "\r\n" );
document.write( "Now we must divide that by 3!=6 because the committees can be permuted in 3!\r\n" );
document.write( "= 6 ways, \r\n" );
document.write( "\r\n" );
document.write( "That's (33C11)(22C11)(11C11)/3! = 22754499243840\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( " \n" ); document.write( "
\n" );