How many simple graphs on a set of 8 vertices have 6 edges?












0














I think the total number of edges for a graph with $8$ vertices would be:



$n(n-1)/2$ which would yield $28$.



total number of set with $28$ elements is $2^{28}$.



But I'm not sure how I can limit the number of edges to be maximum of $6$....










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    0














    I think the total number of edges for a graph with $8$ vertices would be:



    $n(n-1)/2$ which would yield $28$.



    total number of set with $28$ elements is $2^{28}$.



    But I'm not sure how I can limit the number of edges to be maximum of $6$....










    share|cite|improve this question



























      0












      0








      0







      I think the total number of edges for a graph with $8$ vertices would be:



      $n(n-1)/2$ which would yield $28$.



      total number of set with $28$ elements is $2^{28}$.



      But I'm not sure how I can limit the number of edges to be maximum of $6$....










      share|cite|improve this question















      I think the total number of edges for a graph with $8$ vertices would be:



      $n(n-1)/2$ which would yield $28$.



      total number of set with $28$ elements is $2^{28}$.



      But I'm not sure how I can limit the number of edges to be maximum of $6$....







      graph-theory






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      share|cite|improve this question













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      share|cite|improve this question








      edited Nov 25 '13 at 16:37







      user8523

















      asked Nov 25 '13 at 4:57









      user111264

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          HINT: There are indeed $binom82=28$ possible edges. You simply need to pick $6$ of them. How many ways are there to choose $6$ elements from a set of $28$ elements? (This answers the question in your title. If you want the graphs with at most $6$ edges, you’ll have to count the number of ways pick $0,1,2,3,4$, and $5$ of them, as well, and add the results.)






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            1 Answer
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            1 Answer
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            active

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            1














            HINT: There are indeed $binom82=28$ possible edges. You simply need to pick $6$ of them. How many ways are there to choose $6$ elements from a set of $28$ elements? (This answers the question in your title. If you want the graphs with at most $6$ edges, you’ll have to count the number of ways pick $0,1,2,3,4$, and $5$ of them, as well, and add the results.)






            share|cite|improve this answer


























              1














              HINT: There are indeed $binom82=28$ possible edges. You simply need to pick $6$ of them. How many ways are there to choose $6$ elements from a set of $28$ elements? (This answers the question in your title. If you want the graphs with at most $6$ edges, you’ll have to count the number of ways pick $0,1,2,3,4$, and $5$ of them, as well, and add the results.)






              share|cite|improve this answer
























                1












                1








                1






                HINT: There are indeed $binom82=28$ possible edges. You simply need to pick $6$ of them. How many ways are there to choose $6$ elements from a set of $28$ elements? (This answers the question in your title. If you want the graphs with at most $6$ edges, you’ll have to count the number of ways pick $0,1,2,3,4$, and $5$ of them, as well, and add the results.)






                share|cite|improve this answer












                HINT: There are indeed $binom82=28$ possible edges. You simply need to pick $6$ of them. How many ways are there to choose $6$ elements from a set of $28$ elements? (This answers the question in your title. If you want the graphs with at most $6$ edges, you’ll have to count the number of ways pick $0,1,2,3,4$, and $5$ of them, as well, and add the results.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '13 at 5:00









                Brian M. Scott

                455k38505908




                455k38505908






























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