UGC NET CS 2016 July – II

Spanning trees in complete graph is equal to n^(n-2)(where n is no of sides or regularity in complete graph). So, spanning trees in complete graph K₄ will be 4^{(4 - 2)}. i.e. 4² = 16. Spanning trees in a bipartite graph K_m,n is equal to m^(n-1) * n^(m-1). So, spanning trees in K_2,2 will be 2^(2-1) * 2^(2-1). i.e. 2¹ * 2¹.= 4. So, option (C) is correct.

There are BB RR and BR, total outcome is 3, probability that the opposite side is the same color as the one side we observed: It is clear that BB and RR will show the same color on opposite side, so favorable outcome will be 2. Probability will be favorable outcome / total outcome i.e. 2 / 3. So, option (B) is correct.

Total 5 clique will be there. So, option (C) is correct.

We know that x ⊕ y = x`y + xy` So,

1. 1 ⊕ 0 = 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 = 1
2. 1 ⊕ 1 ⊕ 1 = (1 ⊕ 1) ⊕ 1 = (1`1 + 11`) ⊕ 1 = (0) ⊕ 1 = 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 = 1
3. 1 ⊕ 1 ⊕ 0 = 1 ⊕ (1 ⊕ 0 ) = 1 ⊕ ( 1`0 + 10` = 0 * 0 + 1 * 1 = 0 + 1 ) = 1 ⊕ 1 = 1`1 + 11` = 0
4. 1 ⊕ 1 = 1`1 + 11` = 0
So, option (C) is correct.

The IEEE-754 double-precision format to represent floating point numbers, has a length of 64 bits. 1 bit for sign, 11 bit for exponent and rest 52 is for fraction. So, option (D) is correct.

So, option (C) is correct.

(AB` + AB`C + AC) (A`C` + B`) = AB` +AB`C +AB`C = AB` +AB`C = AB`(1 + C) = AB` So, option (A) is correct.