Discrete Mathematics Questions and Answers

Answer: B

P is true and Q is false: \(R\) is reflexive but not transitive.

Reflexive: \((a,b)R(a,b)\) holds because \(a \le a\).

Not transitive: take \((1,5)R(2,1)\), true since \(1 \le 2\), and \((2,1)R(0,3)\), true since \(1 \le 3\). But \((1,5)R(0,3)\) needs \(1 \le 0\) or \(5 \le 3\), and both fail. The “or” lets each step use a different coordinate, so the chain does not carry through.

Hence options calling \(R\) transitive, or not reflexive, are wrong.

Answer: D

The count is \(2\binom{n}{2k}(n-2k)!\,(k!)^{2}\).

Build a separating permutation in steps:

1. Choose the 2k positions that the members of \(A \cup B\) will occupy: \(\binom{n}{2k}\) ways.
2. Fill the other \(n - 2k\) positions with the remaining elements: \((n-2k)!\) ways.
3. Decide which set comes first in the chosen positions: 2 ways. The first k of those positions then hold that set and the last k the other.
4. Arrange A within its k positions and B within its k positions: \(k!\,k! = (k!)^{2}\) ways.

Multiplying gives the answer. The option without the factor 2 counts only one of the two orders, the option without \((k!)^{2}\) forgets to arrange the members inside each set, and \(n!\) counts every permutation, separating or not.