document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1172438>Question 1172438</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Trish has a penny, 3 nickels, 4 dimes, and 3 quarters in her pocket. How many different arrangements are possible if she removes one coin at a time from her pocket? \"46,200\" </SPAN>\n" );
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document.write( "Edit: I'm redoing nearly the entire problem. Thanks ikleyn for pointing out I missed that penny. I'm not sure how I overlooked it.\r<BR>\n" );
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document.write( "Here's one way to approach the problem. We have 1+3+4+3 = 11 coins total. There are 11! = 39,916,800 ways to arrange them. The exclamation mark indicates factorial. We start at 11, and count our way down to 1, multiplying everything.\r<BR>\n" );
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document.write( "So,<BR>\n" );
document.write( "11! = 11*10*9*8*7*6*5*4*3*2*1 = 39,916,800<BR>\n" );
document.write( "This would be the answer if we could tell the nickels apart from other nickels, the dimes apart from the other dimes, and the quarters from the other quarters.\r<BR>\n" );
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document.write( "However, we cannot make such distinctions.\r<BR>\n" );
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document.write( "Since we cannot tell the nickels apart, this means we overcounted by a factor of 3! = 6 just for the nickels portion. Similarly, the dimes are overcounted by a factor of 4! = 24 as well. The quarters were overcounted by a factor of 3! = 6.\r<BR>\n" );
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document.write( "Overall, we overcounted by a factor of 6*24*6 = 864<BR>\n" );
document.write( "We need to divide the result of the 11! by 864 to correct for this erroneous overcount.\r<BR>\n" );
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document.write( "Therefore, there are (11!)/(3!*4!*3!) = (39,916,800)/(864) = 46,200 distinct permutations possible.\r<BR>\n" );
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document.write( "<font color=red>Answer: 46,200</font></font> </SPAN>\n" );
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