Here are solutions for the given problems:

1. Find the number of subsets of $A \times B$:

Given: $A = \{1, 2, 3\}, \quad B = \{1, 4\}$ The Cartesian product $A \times B$ contains $3 \times 2 = 6$ elements: $A \times B = \{(1,1), (1,4), (2,1), (2,4), (3,1), (3,4)\}.$ The number of subsets of a set is $2^n$, where $n$ is the number of elements in the set. Hence: $\text{Number of subsets of } A \times B = 2^6 = 64.$

2. Solve for $a$ and $y$:

Given: $\left( \frac{a}{8} + 1, y - 2|3 \right) = \left( -\frac{2}{3}, 1 | 2|3 \right)$

• Equating the first components: $\frac{a}{8} + 1 = -\frac{2}{3}$ Solve for $a$: $\frac{a}{8} = -\frac{2}{3} - 1 = -\frac{5}{3} \quad \Rightarrow \quad a = -\frac{5}{3} \times 8 = -\frac{40}{3}.$

• Equating the second components: $y - 2|3 = 1|2|3.$ Matching the structure of the expression, $y = 1$.

3. Simplify $\cos(a + y) + \cos(a - y)$:

Using the identity: $\cos(a + y) + \cos(a - y) = 2 \cos(a) \cos(y).$ Thus: $\cos(a + y) + \cos(a - y) = 2 \cos(a) \cos(y).$

4. Solve for $p$:

Given: $p \cos\left(\frac{3\pi}{4} + x\right) + \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2} \cos(x).$ Using trigonometric transformations: $\cos\left(\frac{3\pi}{4} + x\right) = -\frac{\sqrt{2}}{2} \cos(x) - \frac{\sqrt{2}}{2} \sin(x),$ $\cos\left(\frac{3\pi}{4} - x\right) = -\frac{\sqrt{2}}{2} \cos(x) + \frac{\sqrt{2}}{2} \sin(x).$ Add both: $p \left(-\frac{\sqrt{2}}{2} \cos(x) - \frac{\sqrt{2}}{2} \sin(x)\right) + \left(-\frac{\sqrt{2}}{2} \cos(x) + \frac{\sqrt{2}}{2} \sin(x)\right) = -\sqrt{2} \cos(x).$ Simplify for $p$.

5. Solve the inequality:

Given: $\frac{n + 2}{2} + \frac{n}{3} \leq 10 + \frac{n}{6}.$ Combine all terms: $\frac{3(n + 2)}{6} + \frac{2n}{6} \leq 10 + \frac{n}{6}.$ $\frac{3n + 6 + 2n}{6} \leq 10 + \frac{n}{6}.$ $\frac{5n + 6}{6} \leq 10 + \frac{n}{6}.$ Multiply through by 6: $5n + 6 \leq 60 + n.$ Simplify: $5n - n \leq 60 - 6.$ $4n \leq 54 \quad \Rightarrow \quad n \leq \frac{54}{4} = 13.5.$ Solution: $n \in (-\infty, 13.5].$

6. Combinatorics:

(i) $nCr = \frac{n!}{r!(n-r)!}$.

(ii) A committee of 3 men and 2 women chosen from 7 men and 5 women: $\binom{7}{3} \times \binom{5}{2} = \frac{7!}{3!(7-3)!} \times \frac{5!}{2!(5-2)!}.$ Simplify: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35, \quad \binom{5}{2} = \frac{5 \cdot 4}{2 \cdot 1} = 10.$ Total: $35 \times 10 = 350.$

Would you like detailed solutions for any of the steps above?

Follow-up questions:

1. Can you calculate Cartesian products for larger sets?
2. How can trigonometric identities simplify complex expressions?
3. Why is $nCr$ critical in combinatorics?
4. How does solving inequalities differ when fractions are involved?
5. Can the solution of inequalities be represented on graphs?

Tip:

Use symmetry in trigonometric functions like cosine to simplify equations efficiently.