document.write( "Question 345932: A bookshelf holds twelve books in a row. How many ways are there to choose five books, so that no two adjacent books are chosen? \n" ); document.write( "
Algebra.Com's Answer #247743 by Edwin McCravy(20055)\"\" \"About 
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document.write( "We look at the 3rd book we pick.\r\n" );
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document.write( "The books are\r\n" );
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document.write( "1   2   3   4   5   6   7   8   9  10  11  12\r\n" );
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document.write( "You can only pick your 3rd book as 5, 6, 7, or 8, because if you pick your 3rd\r\n" );
document.write( "book closer to the left end you won't be able to pick a 1st and 2nd book and if\r\n" );
document.write( "you pick your 3rd book closer to the right end you won't be able to pick a 4th\r\n" );
document.write( "and 5th book.\r\n" );
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document.write( "By symmetry, there are the same number of ways to pick 5 for your 3rd book as\r\n" );
document.write( "there are ways to pick 8 for your 3rd book.  Similarly there are as many ways\r\n" );
document.write( "to pick 6 for your 3rd book as there are ways to pick 7 for your 3rd book.\r\n" );
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document.write( "So we only need to know: \r\n" );
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document.write( "1. How many ways we can pick 5 for our 3rd book and \r\n" );
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document.write( "2. How many ways we can pick 6 for our 3rd book.\r\n" );
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document.write( "Let's do the first one, where we pick 5 for our 3rd book.\r\n" );
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document.write( "If we pick 5 for our 3rd book, we must pick books 1 and 3 for our 1st and 2nd\r\n" );
document.write( "books.  So we need only to look at the number of ways to pick our 4th and 5th\r\n" );
document.write( "books. \r\n" );
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document.write( "If we pick 7 for our 4th book, we can pick 9-12 for our 5th book.\r\n" );
document.write( "That's 4 ways. \r\n" );
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document.write( "If we pick 8 for our 4th book, we can pick 10-12 for our 5th book.\r\n" );
document.write( "That's 3 more ways.\r\n" );
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document.write( "If we pick 9 for our 4th book, we can pick 11-12 for our 5th book.\r\n" );
document.write( "That's 2 more ways.\r\n" );
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document.write( "If we pick 10 for our 4th book, we can only pick 12 for our 5th book.\r\n" );
document.write( "That's 1 more way.\r\n" );
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document.write( "So there are 4+3+2+1 or 10 ways to pick our 3rd book as 5.\r\n" );
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document.write( "By symmetry there are also 10 ways to pick 8 as our 3rd book.\r\n" );
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document.write( "-----------------------------------------\r\n" );
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document.write( "Let's do the other one, where we pick 6 for our 3rd book.\r\n" );
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document.write( "If we pick 6 for our 3rd book, we can either pick 1 and 3, 1 and 4, or 2 and 4\r\n" );
document.write( "for our 1st and 2nd books.  That's 3 ways to pick our 1st and 2nd books.   \r\n" );
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document.write( "So we need only to look at the number of ways to pick our 4th and 5th books,\r\n" );
document.write( "and multiply 3 by that number.\r\n" );
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document.write( "If we pick 8 for our 4th book, we can pick 10-12 for our 5th book.\r\n" );
document.write( "That's 3 ways. \r\n" );
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document.write( "If we pick 9 for our 4th book, we can pick 11-12 for our 5th book.\r\n" );
document.write( "That's 2 more ways.\r\n" );
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document.write( "If we pick 10 for our 4th book, we can only pick 12 for our 5th book\r\n" );
document.write( "That's 1 more ways.\r\n" );
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document.write( "So there are 3*(3+2+1) or 18 ways to pick our 3rd book as 6.\r\n" );
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document.write( "By symmetry there are also 18 ways to pick 7 for our 3rd book.\r\n" );
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document.write( "Therefore the total number of ways is\r\n" );
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document.write( "10 ways to pick 5 for the 3rd book\r\n" );
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document.write( " plus\r\n" );
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document.write( "18 ways to pick  6 for the 3rd book\r\n" );
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document.write( "plus, by symmetry,\r\n" );
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document.write( "18 ways to pick 7 for the 3rd book\r\n" );
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document.write( "plus\r\n" );
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document.write( "10 ways to pick 8 for the 3rd book\r\n" );
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document.write( "10 + 18 + 18 + 10 = 56\r\n" );
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document.write( "Edwin
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