How many simple graphs on a set of 8 vertices have 6 edges?
$begingroup$
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
asked Nov 25 '13 at 4:57
user111264user111264
12
$endgroup$
add a comment |
$begingroup$
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
asked Nov 25 '13 at 4:57
user111264user111264
12
$endgroup$
add a comment |
$begingroup$
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
asked Nov 25 '13 at 4:57
user111264user111264
12
$endgroup$
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
graph-theory
asked Nov 25 '13 at 4:57
user111264user111264
12
asked Nov 25 '13 at 4:57
user111264user111264
12
edited Nov 25 '13 at 16:37
user8523
asked Nov 25 '13 at 4:57
user111264user111264
12
asked Nov 25 '13 at 4:57
user111264user111264
12
asked Nov 25 '13 at 4:57
user111264user111264
12
12
add a comment |
add a comment |
1 Answer
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$begingroup$
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.)
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
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1 Answer
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1 Answer
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$begingroup$
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.)
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
add a comment |
$begingroup$
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.)
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
add a comment |
$begingroup$
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.)
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
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.)
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
answered Nov 25 '13 at 5:00
Brian M. ScottBrian M. Scott
457k38509909
457k38509909
add a comment |
add a comment |
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