document.write( "Question 1207714: In how many ways can the numbers 1, 2, 3, 4, 5, and 6 be arranged in a row, so that the product of any two adjacent numbers is at least 4? \n" ); document.write( "
Algebra.Com's Answer #845732 by Edwin McCravy(20055)\"\" \"About 
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document.write( "The only possible adjacent positive integer pairs with a \r\n" );
document.write( "product less than 4 are \"12\", \"21\", \"13\", and \"31\".\r\n" );
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document.write( "So we start with 6!=720 arrangements of 1,2,3,4,5,6.\r\n" );
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document.write( "Then we subtract the 5!=120 arrangements of (12),3,4,5,6 and\r\n" );
document.write( "the 5!=120 arrangements of (21),3,4,5,6.\r\n" );
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document.write( "Then we subtract the 5!=120 arrangements of (13),2,4,5,6 and\r\n" );
document.write( "the 5!=120 arrangements of (31),2,4,5,6.\r\n" );
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document.write( "But we have subtracted out those containing \"213\" twice \r\n" );
document.write( "and those containing \"312\" twice.  So we need to add back in\r\n" );
document.write( "the 4!=24 arrangements of (213),4,5,6, and the 4!=24 \r\n" );
document.write( "arrangements of (312),4,5,6, once each.\r\n" );
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document.write( "So the answer is\r\n" );
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document.write( "6! - 4x5! + 2x4! = 288.\r\n" );
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document.write( "Edwin
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