1. **Problem Statement:** We have $P_n = \prod_{k=0}^n C_k^t$, where $C_k$ are the binomial coefficients from $(1+x)^n$. The ratio is given by $$\frac{P_{n+1}}{P_n} = \frac{(n+1)^x}{y!}$$ We need to find $x + y$. 2. **Recall the binomial coefficients:** The binomial coefficients for $(1+x)^n$ are $C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!}$ for $k=0,1,\ldots,n$. 3. **Understanding $P_n$:** $P_n$ is the product of all binomial coefficients for $(1+x)^n$: $$P_n = \prod_{k=0}^n \binom{n}{k}$$ 4. **Formula for product of binomial coefficients:** It is known that $$\prod_{k=0}^n \binom{n}{k} = \frac{(n!)^{n+1}}{\prod_{k=0}^n (k!) (n-k)!}$$ Since $(n-k)!$ runs over the same values as $k!$ in reverse, the denominator is $$\prod_{k=0}^n k! \times \prod_{k=0}^n k! = \left(\prod_{k=0}^n k!\right)^2$$ 5. **Simplify $P_n$:** $$P_n = \frac{(n!)^{n+1}}{\left(\prod_{k=0}^n k!\right)^2}$$ 6. **Compute the ratio $\frac{P_{n+1}}{P_n}$:** $$\frac{P_{n+1}}{P_n} = \frac{\frac{((n+1)!)^{n+2}}{\left(\prod_{k=0}^{n+1} k!\right)^2}}{\frac{(n!)^{n+1}}{\left(\prod_{k=0}^n k!\right)^2}} = \frac{((n+1)!)^{n+2}}{(n!)^{n+1}} \times \frac{\left(\prod_{k=0}^n k!\right)^2}{\left(\prod_{k=0}^{n+1} k!\right)^2}$$ 7. **Rewrite factorial products:** Note that $$\prod_{k=0}^{n+1} k! = \left(\prod_{k=0}^n k!\right) \times (n+1)!$$ So, $$\frac{\left(\prod_{k=0}^n k!\right)^2}{\left(\prod_{k=0}^{n+1} k!\right)^2} = \frac{\left(\prod_{k=0}^n k!\right)^2}{\left(\prod_{k=0}^n k!\right)^2 ((n+1)!)^2} = \frac{1}{((n+1)!)^2}$$ 8. **Substitute back:** $$\frac{P_{n+1}}{P_n} = \frac{((n+1)!)^{n+2}}{(n!)^{n+1}} \times \frac{1}{((n+1)!)^2} = \frac{((n+1)!)^{n+2-2}}{(n!)^{n+1}} = \frac{((n+1)!)^{n}}{(n!)^{n+1}}$$ 9. **Express factorials:** Recall $(n+1)! = (n+1) \times n!$, so $$\frac{P_{n+1}}{P_n} = \frac{((n+1) n!)^n}{(n!)^{n+1}} = \frac{(n+1)^n (n!)^n}{(n!)^{n+1}} = \frac{(n+1)^n}{n!}$$ 10. **Compare with given ratio:** Given $$\frac{P_{n+1}}{P_n} = \frac{(n+1)^x}{y!}$$ From our result, $$\frac{P_{n+1}}{P_n} = \frac{(n+1)^n}{n!}$$ 11. **Identify $x$ and $y$:** $$x = n, \quad y = n$$ 12. **Find $x + y$:** $$x + y = n + n = 2n$$ **Answer:** $2n$ **Option:** B) 2n