document.write( "Question 610605: 10 people are to be divided into 3 committees, in such a way that every committee must have at least one member. and no person can serve on all three committees. In how many ways can this be done. \n" ); document.write( "
Algebra.Com's Answer #384485 by AnlytcPhil(1806)\"\" \"About 
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document.write( "There are 8 ways to divide into 3 committees: \r\n" );
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document.write( "   1st        2nd        3rd\r\n" );
document.write( "committee  committee  committee\r\n" );
document.write( " has this   has this   has this \r\n" );
document.write( "  many       many       many\r\n" );
document.write( " members    members    members\r\n" );
document.write( "    1          1          8      C(10,1)·C(9,1)·C(8,8)/2 = 10·9·1/2 = 45 ways  \r\n" );
document.write( "    1          2          7      C(10,1)·C(9,2)·C(7,7)=10·36·1 = 360 ways  \r\n" );
document.write( "    1          3          6      C(10,1)·C(9,3)·C(6,6)=10·84·1 = 840 ways\r\n" );
document.write( "    1          4          5      C(10,1)·C(9,4)·C(5,5)=10·126·1 = 1260 ways\r\n" );
document.write( "    2          2          6      C(10,2)·C(8,2)·C(6,6)/2 = 45·28·1/2 = 630 ways\r\n" );
document.write( "    2          3          5      C(10,2)·C(8,3)·C(5,5) = 45·56·1 = 560 ways\r\n" );
document.write( "    2          4          4      C(10,2)·C(8,4)·C(4,4) = 45·70·1/2 = 1575 ways\r\n" );
document.write( "    3          3          4      C(10,3)·C(7,3)·C(4,4) = 120·35·1/2 = 2100 ways\r\n" );
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document.write( "Note: In the four cases where there are two committees of the same size,\r\n" );
document.write( "we must divide by 2 in order to avoid ordering those two committees.\r\n" );
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document.write( "Total = 45+360+840+1260+630+560+1575+2100 = 7370 ways.\r\n" );
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document.write( "Edwin

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