i The graph Ks is bipartite G If we choose a permutation of

(i) The graph Ks is bipartite. (G) If we choose a permutation of (1,.., 7) at random, there is a probability of exactly that the number 1 comes to the left of the number 7.

Solution

i)

False

For this we must be able to split the graph into two sets say, A and B so that no two vertices in A have an edge in common. And no two vertices in B have an edge in common. But K3 has 3 vertices so either A or B has 2 vertices

Assume,A has 2 vertices. But k3 is complete graph so they have edge between them. Hence graph is not bipartite.

j)

True

No two numbers can occupy the same place. Hence 1 can either come to the left of number 7 or to the right of number 7

But since there is not preference or bias to any of the same number so number of permutations where 1 comes to left of 7 is equal to number of permutations where 1 comes to right of 7

Hence proved.