document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1114788>Question 1114788</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello, thank you for helping me with this math question<BR>\n" );
document.write( "Consider a class with 12 girls and 10 boys (22 in total):<BR>\n" );
document.write( "In how many ways can a committee of 5 consisting of 3 girls and 2 boys be chosen?<BR>\n" );
document.write( "a) 0.03<BR>\n" );
document.write( "b) 792<BR>\n" );
document.write( "c) 0.3759<BR>\n" );
document.write( "d) 9900<BR>\n" );
document.write( "Here is my thought process which led to none of the given answer choices.<BR>\n" );
document.write( "We can easily eliminate A and C, as they are not numbers of different arrangements, rather, they are probabilities, which is not what the question is asking.<BR>\n" );
document.write( "I recognized the nature of the question to be a permutation, as you care about which order you are placing the 12 girls in 3 places on the committee. For example, Girl A, B, C is a possible permutation, but B, A, C is also a different permutation that must be counted. The same is true for putting the 10 boys into the 2 other spaces on the committee. So, I tried (12 nPr 3) + (10 nPr 2), which made logical sense in my head as you want 12 girls in 3 different ways on the committee plus the 10 boys you want in 3 different ways on the committee.<BR>\n" );
document.write( "This way gets you to an answer of 1410, which is way off from the answer choices,so I was wondering what I did wrong in this thought process.<BR>\n" );
document.write( "Thank you for your time and free advice :)<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #729703 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=729703\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "(1) It is NOT a permutation; on a committee the order is NOT important.<br><BR>\n" );
document.write( "(2) For each of the ways of choosing 3 of the 12 girls there are many ways of choosing 2 of the 10 boys.  That \"for each\" means you need to multiply the two numbers together -- not add them.<br><BR>\n" );
document.write( "Fix those two parts of your analysis of the problem and you should reach answer d. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );