document.write( "Question 1191220: 6. From 11 novels and 3 dictionaries, 4 novels and 1 dictionary are to be selected and arranged on a shelf so that the dictionary is always in the middle. How many such arrangements are possible?\r
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Algebra.Com's Answer #823015 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "The wording of \"middle\" is unfortunately vague. \r
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\n" ); document.write( "\n" ); document.write( "If your teacher means the exact middle, then we'd have 2 novels on one side of the dictionary and 2 novels on the other side.\r
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\n" ); document.write( "\n" ); document.write( "Concerning the novels only for now, there are 4 slots to fill and 11 novels to pick from.\r
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\n" ); document.write( "\n" ); document.write( "Use n = 11 and r = 4 to plug into the nPr formula.
\n" ); document.write( "n P r = (n!)/( (n-r)! )
\n" ); document.write( "11 P 4 = (11!)/( (11-4)! )
\n" ); document.write( "11 P 4 = (11!)/( 7! )
\n" ); document.write( "11 P 4 = (11*10*9*8*7!)/( 7! )
\n" ); document.write( "11 P 4 = 11*10*9*8
\n" ); document.write( "11 P 4 = 7920
\n" ); document.write( "There are 7920 ways to arrange the 4 novels from a total collection of 11 novels.
\n" ); document.write( "This is if your teacher considers the order mattering.
\n" ); document.write( "This applies to one dictionary. But we have 3 of them, so 3*7920 = 23,760\r
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\n" ); document.write( "\n" ); document.write( "Answer: 23,760
\n" ); document.write( "This of course only applies if the dictionary is right in the exact middle. Also, it involves the order mattering.
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