1. Problem: Calculate the value of $\frac{8!}{2! \times 6!}$. Formula: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, here $\frac{8!}{2!6!}$ is a combination $\binom{8}{2}$. Calculation: $$\frac{8!}{2!6!} = \frac{8 \times 7 \times 6!}{2! \times 6!} = \frac{8 \times 7}{2 \times 1} = 28$$ Answer: A. 28 2. Problem: Calculate $C^4_1 \times C^4_3$. Formula: $C^n_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$. Calculate each: $$C^4_1 = \binom{4}{1} = 4$$ $$C^4_3 = \binom{4}{3} = 4$$ Multiply: $$4 \times 4 = 16$$ Answer: C. 16 3. Problem: Calculate $P^5_2 \times P^5_4$. Formula: $P^n_r = \frac{n!}{(n-r)!}$. Calculate each: $$P^5_2 = \frac{5!}{(5-2)!} = \frac{120}{6} = 20$$ $$P^5_4 = \frac{5!}{(5-4)!} = \frac{120}{1} = 120$$ Multiply: $$20 \times 120 = 2400$$ Since options are much smaller, re-check problem: Possibly a typo or intended to multiply permutations but options suggest smaller number. If question means product of permutations counts, answer is 2400 (not in options). 4. Problem: Simplify $\frac{(n+1)!}{(n-1)!}$. Rewrite numerator: $$(n+1)! = (n+1) \times n \times (n-1)!$$ Divide: $$\frac{(n+1)!}{(n-1)!} = (n+1) \times n \times \frac{(n-1)!}{(n-1)!} = n(n+1) = n^2 + n$$ Answer: None of the options exactly match $n^2 + n$, but closest is E. $n^2 + n + 1$ (extra +1). Check options again: A. $n+1$ B. $n$ C. $n^2 - 1$ D. $n^2 + 1$ E. $n^2 + n + 1$ Correct simplified form is $n^2 + n$, which is not exactly listed, so answer is none of the above. But since $n^2 + n$ is closest to E, possibly a typo. Final answer: $n^2 + n$ (not exactly in options). Summary: 1. Answer: 28 2. Answer: 16 3. Answer: 2400 (not in options) 4. Simplified form: $n^2 + n$