document.write( "Question 1154710: I have five children and want to place them in a line for a photograph.
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Algebra.Com's Answer #777184 by Edwin McCravy(20055) $\"\"$ $\"About$

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Let the 5 children be Hugh, Louise, Richard, Ann, and Bob.\r\n" );
document.write( "Let their initials stand for them.\r\n" );
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document.write( "There would be 5! = 5∙4∙3∙2∙1 = 120 ways if there were no restrictions.\r\n" );
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document.write( "Let's enumerate the unwanted cases where H is between L and R, and subtract\r\n" );
document.write( "them from the 120.\r\n" );
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document.write( "There are 2 ways to arrange them with Hugh between them. We can have _L_H_R_\r\n" );
document.write( "or _R_H_L_, where the blanks can contain 0, 1 or 2 people. \r\n" );
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document.write( "For each of those 2 ways, there are two cases:\r\n" );
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document.write( "Case 1: A and B are not side by side.\r\n" );
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document.write( "There are 4 blanks to put A in, and 3 remaining blanks to put B in.\r\n" );
document.write( "That's 4∙3=12 for Case 1.\r\n" );
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document.write( "Case 2: A and B are side-by-side\r\n" );
document.write( "They can be side-by-side 2 ways, AB or BA\r\n" );
document.write( "For either of those 2 ways, there are 4 places to put them.\r\n" );
document.write( "That's 2∙4=8 ways for Case 2.\r\n" );
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document.write( "The total for those two cases is 12+8=20.\r\n" );
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document.write( "So the number of unwanted arrangements is 2∙(12+8)=2∙20 = 40 unwanted cases.\r\n" );
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document.write( "We subtract that from the 120 and so the answer is 120-40 = 80.\r\n" );
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document.write( "Edwin

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