document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170761>Question 1170761</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A school dance committee is to consist of 2 freshmen, 3 sophomores, 4 juniors, and 5 seniors. If 7 freshmen, 9 sophomores, 9 juniors, and 9 seniors are eligible to be on the committee, in how many ways can the committee be chosen? <BR>\n" );
document.write( "Your answer is :    <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #795642 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=795642\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "We have 7 freshmen total, and we want to pick 2 of them. There are 7*6 = 42 ways to do this if order mattered; however it doesn't. So we divide by 2 to get 42/2 = 21. We could use the nCr combination formula with n = 7 and r = 2.\r<BR>\n" );
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document.write( "There are 9 sophomores and 3 spots for them. There are 9*8*7 = 504 permutations and 504/(3!) = 504/(3*2*1) = 504/6 = 84 combinations. We divide by 3! = 6 because there are 6 ways to arrange any group of three people.\r<BR>\n" );
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document.write( "There are 9 juniors and 4 slots for them. Let's use the nCr combination formula with n = 9 and r = 4 and we get<BR>\n" );
document.write( "nCr = (n!)/(r!*(n-r)!)<BR>\n" );
document.write( "9C4 = (9!)/(4!*(9-4)!)<BR>\n" );
document.write( "9C4 = (9!)/(4!*5!)<BR>\n" );
document.write( "9C4 = (9*8*7*6*5!)/(4!*5!)<BR>\n" );
document.write( "9C4 = (9*8*7*6)/(4!) a pair of \"5!\" terms cancel<BR>\n" );
document.write( "9C4 = (9*8*7*6)/(4*3*2*1)<BR>\n" );
document.write( "9C4 = 3024/24<BR>\n" );
document.write( "9C4 = 126<BR>\n" );
document.write( "There are 126 ways to pick the four juniors from a total of nine.\r<BR>\n" );
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document.write( "Finally, there are 9 seniors and 5 seats for them on the committee. Plug in n = 9 and r = 5<BR>\n" );
document.write( "nCr = (n!)/(r!*(n-r)!)<BR>\n" );
document.write( "9C5 = (9!)/(5!*(9-4)!)<BR>\n" );
document.write( "9C5 = (9!)/(5!*5!)<BR>\n" );
document.write( "9C5 = (9*8*7*6*5!)/(5!*4!)<BR>\n" );
document.write( "9C5 = (9*8*7*6)/(4!)<BR>\n" );
document.write( "9C5 = (9*8*7*6)/(4*3*2*1)<BR>\n" );
document.write( "9C5 = 3024/24<BR>\n" );
document.write( "9C5 = 126<BR>\n" );
document.write( "There are 126 ways to pick the five seniors from a total of nine.<BR>\n" );
document.write( "We get the same result because of the symmetry found in the combination formula. \r<BR>\n" );
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document.write( "You could use Pascal's triangle to get this value, or any other value found earlier. In this case, look in the row that starts with \"1,9,...\". Start the counter at 0 and count up until you get to r = 5. Each time you increase the counter, you move to the right 1 spot. Following this process should lead you to 126.\r<BR>\n" );
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document.write( "So we have found<BR>\n" );
document.write( "21 ways to pick the freshmen<BR>\n" );
document.write( "84 ways to pick the sophomores<BR>\n" );
document.write( "126 ways to pick the juniors<BR>\n" );
document.write( "126 ways to pick the seniors\r<BR>\n" );
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document.write( "Multiply those values out: 21*84*126*126 = 28,005,264\r<BR>\n" );
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document.write( "Answer:  28,005,264 different committees.\r<BR>\n" );
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