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The number of labelled subgraphs possible for the graph given below.

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112 is the answer for k4.. the solution to the given question in the mock test is wrong
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check the solution given below..don't know any formulae..i just follow the procedure.
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1 Answer

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One vertex sub-graphs: 4C1 (number of ways of picking one vertex among the four)

Two vertices sub-graphs: 4*2^1 (whether to include the edge available between the given selected set of vertices) + 2 * 1 as there is no edge between these set of vertices.

Three vertices sub-graphs: 4C3 * 2^2 (two edges incident on the any selected 3 vertices, 2^2 different ways of having/not-having these edges in the sub-grah)

Four vertices sub-graphs: 4C4 * 2^4 ( four edges)

Total number of subgraphs: 4 + 10 + 4C3 * 2^2 + 4C4 * 2^4 = 46

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10 Comments

Bro, u commented as 112 now posted answer as 48?

:(
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sorry..112 was for k4..mistyped it
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@balchandar reddy san

"Two vertices sub-graphs: 4C2 * 2 (whether to include the edge available between the given selected set of vertices)"

what if the 2 out of 4 vertices, b and d are selected and there is no edge between b and d, how can you consider the possibility of selecting and edge?? :/

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he is correct 112 will be  for k4 and if graph will be  changed than no fixed formula to that only we can apply counting
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@chauhansunil20th Yes..you are right, in that case it will be 4*2 + 2*1 which will be 10 for Two-vertex sub-graphs...updated the answer

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nice catch bro. very fine point .
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What about null graph?

Should we have to consider it or not?
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As in the question we are asked to find the number of labelled sub-graphs, so i assumed null graph won't be included.
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no empty graph will not be consider. it is not only in labeled also in unlabebeled. i think u want to ask empty graph. null graph have no edges and empty graph have no vertex and no edges also.
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bro u consider null graph in ur counting like if u take 4C0 and no edges than that was null graph .
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