document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1196471>Question 1196471</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A team of 6 people is to be chosen from 10 people. Two of the people are sisters who must be separated. Find the number of different teams that can be formed. </SPAN>\n" );
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document.write( "Let's say the sisters have codenames of A and B.\r<BR>\n" );
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document.write( "For now, let's have sister A be on the team.<BR>\n" );
document.write( "That means there are 6-1 = 5 seats left to be filled.<BR>\n" );
document.write( "There are 10-2 = 8 people left to fill them. <BR>\n" );
document.write( "Sister B cannot be on the team in this case.\r<BR>\n" );
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document.write( "There are n = 8 people to pick from, and r = 5 seats to fill<BR>\n" );
document.write( "Use the nCr combination formula. <BR>\n" );
document.write( "n C r = (n!)/(r!(n-r)!)<BR>\n" );
document.write( "8 C 5 = (8!)/(5!*(8-5)!)<BR>\n" );
document.write( "8 C 5 = (8!)/(5!*3!)<BR>\n" );
document.write( "8 C 5 = (8*7*6*5!)/(5!*3!)<BR>\n" );
document.write( "8 C 5 = (8*7*6)/(3!)<BR>\n" );
document.write( "8 C 5 = (8*7*6)/(3*2*1)<BR>\n" );
document.write( "8 C 5 = (336)/(6)<BR>\n" );
document.write( "8 C 5 = 56<BR>\n" );
document.write( "There are 56 combinations possible where sister A is on the team, but sister B is not.\r<BR>\n" );
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document.write( "Side note: you can use Pascal's Triangle as an alternative to the nCr formula.\r<BR>\n" );
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document.write( "The calculations will be identical if sister B was on the team, with sister A off the team. <BR>\n" );
document.write( "So we have 56 ways to do this scenario.\r<BR>\n" );
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document.write( "That's 56+56 = 112 combinations so far.\r<BR>\n" );
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document.write( "Now consider the scenario where both sisters are off the team. <BR>\n" );
document.write( "There are n = 10-2 = 8 people to pick from and r = 6 seats to fill. <BR>\n" );
document.write( "Apply the nCr combination formula again.  <BR>\n" );
document.write( "n C r = (n!)/(r!(n-r)!)<BR>\n" );
document.write( "8 C 6 = (8!)/(6!*(8-6)!)<BR>\n" );
document.write( "8 C 6 = (8!)/(6!*2!)<BR>\n" );
document.write( "8 C 6 = (8*7*6!)/(6!*2!)<BR>\n" );
document.write( "8 C 6 = (8*7)/(2!)<BR>\n" );
document.write( "8 C 6 = (8*7)/(2*1)<BR>\n" );
document.write( "8 C 6 = (56)/(2)<BR>\n" );
document.write( "8 C 6 = 28<BR>\n" );
document.write( "There are 28 ways to have a team where both sisters are off the team. \r<BR>\n" );
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document.write( "In total we have 112+28 = 140 different teams<BR>\n" );
document.write( "For any given team, the order doesn't matter. This is why we go for nCr instead of nPr.\r<BR>\n" );
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document.write( "Another approach:\r<BR>\n" );
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document.write( "This may seem counter-intuitive, but let's have the sisters be on the team together for now. <BR>\n" );
document.write( "There are r = 6-2 = 4 seats left to pick with n = 10-2 = 8 people to pick from\r<BR>\n" );
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document.write( "Use the nCr combination formula to get 8C4 = 70<BR>\n" );
document.write( "There are 70 teams with the two sisters on the team together.\r<BR>\n" );
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document.write( "There are 10C6 = 210 possible teams whether we have one sister, both sisters, or no sisters on the team.\r<BR>\n" );
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document.write( "That must mean there are 210-70 = 140 teams with at most one sister. Either there's one sister or no sisters on the team.\r<BR>\n" );
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document.write( "Answer: 140<BR>\n" );
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