document.write( "Question 1151217: There are 9 people lining up today to pay telephone bills. If three persons don:t want to follow each other, how many possible line-ups is possible? \n" ); document.write( "

Algebra.Com's Answer #772889 by ikleyn(52786) $\"\"$ $\"About$

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document.write( "The total number of all possible line-ups is equal to the number of all permutations of 9 objects. i.e. 9! = 9*8*7*6*5*4*3*2*1.\r\n" );
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document.write( "From this number, you should subtract the number of all line-ups, where these 3 persons follow each other.\r\n" );
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document.write( "The number of such line-ups is  7*6!*3!= 7*(6*5*4*3*2*1)*(3*2*1).\r\n" );
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document.write( "The very first factor of 7 accounts for any of 7 possible positions of this triple in the line.\r\n" );
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document.write( "The factor 6! accounts for all possible permutations of remaining 9-3 = 6 persons.\r\n" );
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document.write( "Finally, the factor 3! accounts for all internal permutations inside this triple.\r\n" );
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document.write( "Thus the final answer is\r\n" );
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document.write( "    9*8*7*6*5*4*3*2*1 - 7*(6*5*4*3*2*1)*(3*2*1) = 332640.    ANSWER\r\n" );
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