document.write( "Question 62971: Richard Stanley writes in his book \"Enumerative Combinatorics\" that it is easily seen why \"f%28n%29+=+f%28n-1%29%2Bf%28n-2%29\" where f(n) is the number of subsets of [n] ({1,2,3,..,n}) that do not contain two consecutive integers.\r
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\n" ); document.write( "\n" ); document.write( "Can someone actually explain why it is so, please?\r
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\n" ); document.write( "\n" ); document.write( "Thanks!
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Algebra.Com's Answer #43945 by joyofmath(189)\"\" \"About 
You can put this solution on YOUR website!
Richard Stanley writes in his book \"Enumerative Combinatorics\" that it is easily seen why \"f%28n%29+=+f%28n-1%29%2Bf%28n-2%29\" where f(n) is the number of subsets of [n] ({1,2,3,..,n}) that do not contain two consecutive integers.\r
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\n" ); document.write( "\n" ); document.write( "Can someone actually explain why it is so, please?
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\n" ); document.write( "\n" ); document.write( "This is an interesting problem that took me a while to figure out. I can give you the gist of what's going on but you'd have to do a proof, either by induction or by contradiction to really prove the relationship.\r
\n" ); document.write( "\n" ); document.write( "Here's the idea:\r
\n" ); document.write( "\n" ); document.write( "Let's enumerate the subsets when n=3: {}{1}{2}{3}{13}
\n" ); document.write( "Let's enumerate the subsets when n=4: {}{1}{2}{3}{4}{13}{14}{24}\r
\n" ); document.write( "\n" ); document.write( "Now, let's say we want to construct the subsets when n=5. We can do this as follows, in two steps:\r
\n" ); document.write( "\n" ); document.write( "1. Take all the subsets we used for n=4. They're all still legal when n=5.
\n" ); document.write( "2. Take all the subsets we used for n=3. Add the number 5 to the end of each subset. That would give us {5}{15}{25}{35}{135}.\r
\n" ); document.write( "\n" ); document.write( "Combining the subsets from n=3 (with the 5 tacked onto the end of each subset) and n=4 gives us the subsets for n=5 which is the motivation behind \"f%28n%29+=+f%28n-1%29%2Bf%28n-2%29\".\r
\n" ); document.write( "\n" ); document.write( "Note that there's no duplication between the subsets in n=3 with the 5's tacked on and the subsets in n=4 because the subsets in n=4 can't have any 5's (because n=4). Note also that tacking 5 onto the subsets in n=3 is legal since the n=3 subsets all end in 3 or less and thus tacking on 5 doesn't violate the nonconsecutive number constraint.\r
\n" ); document.write( "\n" ); document.write( "Does this make sense?\r
\n" ); document.write( "\n" ); document.write( "Note, you'd have to do a real proof, in particular you'd need to prove that no subsets were missing when we construct f(n) from f(n-1) and f(n-2).
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