document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1085775>Question 1085775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi, I just want to clarify something:<BR>\n" );
document.write( "Is there a rule for factorials that is a!-(b!c!) = ac!-bc! ?\r<BR>\n" );
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document.write( "An example is 8! - (7!2!), which apparently simplifies to 8*7! - 2*7! = 6*7! = 30240\r<BR>\n" );
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document.write( "Could you please explain how this 'formula' works?\r<BR>\n" );
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document.write( "Thank you very much </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #699865 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=699865\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There is no  general rule for subtraction and/or addition of factorials (none that I know of).   Most of the general rules with factorials involve division and multiplication.  \r<BR>\n" );
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document.write( "We CAN write:<BR>\n" );
document.write( "a! - (b!c!) =  b! * (  a!/b! - c! )\r<BR>\n" );
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document.write( "[ This is no different than  x + yz =   y( x/y + z )  where we factor out a y from each term but since x may or may not have a common factor we must show it as  x/y inside the paren's.  In factorials, there WILL  be common factors so long as b less than a, but I think that's the most we can generally say. ]<BR>\n" );
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document.write( "For your example case:   8! - (7!2!)  = 7! ( 8!/7! - 2! ) = 7!( 8 - 2 ) =  7! * 6<BR>\n" );
document.write( "8!/7! simplifies to 8 because 8! / 7!  = (8*7*6*…*3*2) / (7*6*…*3*2) and all but the 8 cancel.<BR>\n" );
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document.write( "OR, equivalently (more clearly?):   8! - (7!2!) = 8*7! - 2!7! =  7!(8 - 2!) = 7!(8-2) = 7! * 6<BR>\n" );
document.write( "In this approach, we first rewrote 8! as 8*7! then factored out the common 7!.  <BR>\n" );
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document.write( "Edit:  I said there will be common factors in  b! *(a!/b! - c!)  \"so long as b < a\" but I should have said there will \"ALWAYS be common factors between a! and b!\"  (its just if b > a you end up with 1/k  where k>1).  \r<BR>\n" );
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