As m - m = 0 and 0 = 0 \(\times\)12, hence mRm proving that R is reflexive
Let mRn; Then m -n = 12k for some integer k, thus n - m = 12 (-k) and hence nRm. This shows that R is symmetric.
Let mRn and nRp; then m - n =12k and n - p = 12l for some integers k and l.
So m - p = 12 (k + l) and hence mRp. This shows that R is transitive.
Thus R is an equivalence relation.

We have \(n(A \cup B)=6,n(A\cap B)=2\) and \(n(A \triangle B)=4.\)
So, \(n((A\cup B)\times(A\cap B)\times(A \triangle B))=n(A \cup B)\times n(A\cap B)\times n(A\triangle B)= 6 \times 2 \times 4 = 48.\)

Let A and B be the two sets with n(A) = m and n(B) = k. Since A contains more elements than B, we have m > k. From the given conditions we see that 2^m - 2^k = 112. Thus we get, 2k (2^m-k - 1) = 2⁴ \(\times\) 7.
Then the only possibility is k = 4 and 2^{m -k} - 1 = 7. So m - k = 3 and hence m = 7.

For every subset C of {6, 7, 8, 9, 10}, let B = C \(\cup \) {1, 2, 3, 5}. Then A - B = {4}. In other words, for every subset C of {6, 7, 8, 9, 10}, we have a unique set B so that A - B = {4}.
So number of sets \(B \subseteq X\) such that A - B = {4} and the number of subsets of {6, 7, 8, 9, 10} are the same.
So the number of sets \(B \subseteq X\) such that A - B = {4} is 2⁵ = 32.

Let l, m, n ∈ p.
Reflexivity: We cannot say l is perpendicular to l itself.
∴ l R l \(\Rightarrow \) R  is not reflexive.

Symmetry:  lRm ≠ mRl
I is perpendicular to m ⇒ m is perpendicular to l
∴  R is symmetric
Transitive: lRm and mRn ≠ lRn.
l is perpendicular to m and m is perpendicular to n.
⇒ l is perpendicular to n.
∴ R is not transitive.
⇒ R is only symmetric.

n(A\(\times\) A) = 16. \(\Rightarrow\) n(A) = 4
Given S = {a,b) \(\in \) A\(\times\)A : a
\(\therefore\) A = {0, 1, 2, -1}
(A\(\times\)A) = {(0, 0)(0, 1)(0, 2)(0, -1),(1, 1) (1, 2)(1, -1)(2, 0)(2, 1)(2, 2)(2, -1)(-1, 0)(-1, 1)(-1, 2)(-1, -1)}
Now, S = {(0, 1)(0, 2)(-1, 0)(-1, 1)(-1, 2)}
\(\therefore\) The remaining elements of S are (0, 2)(-1, 0)(-1, 1)(1, 2)(0, 1) (1, 2)(0, 2)

Given A = {a, b,c}
(i) Let R = {(a, a) (b, b) (c,c)}
R is reflexive
R is symmetric and R is transitive \(\Rightarrow\) R is an equivalence relation.
This is the equivalence relation of smallest cardinality on A.
\(\therefore\) n(R) = 3
(ii) Let R = {(a, a) (a, b) (a,c) (b,a) (b, b) (b, c) (c, a) (c, b) (c,c)}
R is reflexive since (a, a) (b, b) and (c, c) \(\in \) R
R is symmetric since (a, b) \(\in \) R \(\Rightarrow\)(b, a) \(\in \) R
(b, c) \(\in \) R \(\Rightarrow\) (c, a) \(\in \) R
(c, a) \(\in \) R \(\Rightarrow\) (a, c) \(\in \)R
R is also transitive since (a, b) (b, c) \(\in \) R \(\Rightarrow\) (a, c) \(\in \)R
Hence R is are equivalence relation of largest cardinality on A.
\(\therefore\) n(R) = 9

Since \(P(\phi)\) contains 1 element, \(P(P(\phi))\) contains 2¹ elements and hence \(P(P(P(\phi)))\) contains 2² elements. That is, 4 elements.

\((A\cap B)'\cap A=(A'\cup B')\cap A=(A'\cap A)\cup(B'\cap A)\)
\(=\oslash\cup(B'\cap A)=(B'\cap A)\)
= A - B
So \(n((A\cap B)'\cap A)=n(A-B))=n(A)-n(A\cap B)=7\)

To find n(P(A)), we need n(A)
Let \(n(A\cap B)=k\). Then \(n(A-B)=2k\) and \(n(B-A)=4k\)
Now \(n(A \cup B)=n(A-B)+n(B-A)+n(A\cap B)=7k\).
It is given that \(n(A \cup B)=14\). Thus 7k = 14 and hence k = 2
So n(A - B) = 4 and n(B - A) = 8. As \(n(A) = n(A - B) + n(A \cap B),\) we get n(A) = 6 and hence n(P(A)) = 2⁶ = 64