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Glorfindel
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If you consider isomorphic graphs different, then obviously the answer is $2^{n\choose 2}$. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically $2^{n\choose 2}/n!$. This goes back to a famous method of Pólya (1937), see this paperthis paper for more information. You can find Pólya's original paper herehere.

If you consider isomorphic graphs different, then obviously the answer is $2^{n\choose 2}$. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically $2^{n\choose 2}/n!$. This goes back to a famous method of Pólya (1937), see this paper for more information. You can find Pólya's original paper here.

If you consider isomorphic graphs different, then obviously the answer is $2^{n\choose 2}$. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically $2^{n\choose 2}/n!$. This goes back to a famous method of Pólya (1937), see this paper for more information. You can find Pólya's original paper here.

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GH from MO
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If you consider isomorphic graphs different, then obviously the answer is $2^{n\choose 2}$. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically $2^{n\choose 2}/n!$. This goes back to a famous method of Pólya (1937), see this paper for more information. You can find Pólya's original paper here.

$2^{n\choose 2}$

If you consider isomorphic graphs different, then obviously the answer is $2^{n\choose 2}$. Most graphs have no nontrivial automorphisms, so up to isomorphism the number of different graphs is asymptotically $2^{n\choose 2}/n!$. This goes back to a famous method of Pólya (1937), see this paper for more information. You can find Pólya's original paper here.

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GH from MO
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$2^{n\choose 2}$