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5 Answers

Best answer
12 votes
12 votes
Forest is nothing but disjoint union of trees.

So no of edges in 'k' components of 'v' vertices is nothing but $\sum_{i=1}^k v_i -1$,

where $v_i$ is the no. of vertices in the $i^{th}$ component.

$\sum_{i=1}^k v_i = v$.

So, $\sum_{i=1}^k v_i - 1 = v - k$
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1 votes
1 votes
if n1,n2.... nk are k components.since all k componets are tree then the edges in ni component is (ni-1). So total Edges=

(n1+n2+...nk)-k=v-k.So the option is C
0 votes
0 votes
For a tree the rule is : E = N-1 (where E = no of edges and N = Number of edges)
For a forest that is nothing but lots of disconnected trees this will then look like (E1+E2+E3..... = ((N1-1)+(N2-1)+(N3-1).....untill all the k-components are over)) So for k-components it becomes (E=N-K) here (E= v-k)
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