Key Concepts and Formulas

• Combination Formula: ${ }^n C_r = \frac{n!}{r!(n-r)!}$
• Pascal's Identity: ${ }^n C_r + { }^n C_{r+1} = { }^{n+1} C_{r+1}$
• Permutation Formula: ${ }^n P_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)...(n-r+1)$

Step-by-Step Solution

Step 1: Simplify the first inequality using Pascal's Identity.

We are given ${ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$. We can rewrite the left side as: ${ }^6 C_m + { }^6 C_{m+1} + { }^6 C_{m+1} + { }^6 C_{m+2} > { }^8 C_3$ Applying Pascal's Identity to the first two terms, we get: ${ }^7 C_{m+1} + { }^6 C_{m+1} + { }^6 C_{m+2} > { }^8 C_3$ Applying Pascal's Identity again to the last two terms, we get: ${ }^7 C_{m+1} + { }^7 C_{m+2} > { }^8 C_3$ Applying Pascal's Identity one last time, we have: ${ }^8 C_{m+2} > { }^8 C_3$

Step 2: Determine the value of m.

Since ${ }^8 C_{m+2} > { }^8 C_3$, we need to consider the properties of binomial coefficients. The binomial coefficients ${ }^n C_r$ increase until the middle term(s) and then decrease. So, ${ }^8 C_x$ is greater than ${ }^8 C_3$ if x is closer to 4 (the middle term) than 3, or if x is further from 4 on the other side. If $m+2 = 2$ then $m = 0$ and $^8C_2 = \frac{8 \cdot 7}{2} = 28$. $^8C_3 = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$. In this case $^8C_2 < ^8C_3$ so $m$ cannot be 0. If $m+2 = 4,5,6,7,8$ then $^8C_{m+2} < ^8C_3$ does not hold. Since $^8C_x = ^8C_{8-x}$, we have $^8C_3 = ^8C_5$. Therefore, we need to have $m+2 < 3$ or $m+2 > 5$.

We are looking for an m such that $^8C_{m+2} > ^8C_3 = 56$. If $m+2 = 0, 1, 2, 3$ we have $^8C_0 = 1$, $^8C_1=8$, $^8C_2 = 28$, $^8C_3 = 56$. If $m+2 = 5, 6, 7, 8$ we have $^8C_5 = 56$, $^8C_6 = 28$, $^8C_7=8$, $^8C_8 = 1$.

So, $m+2$ must be either $4$. That is $^8C_4 = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70 > 56$. So $m+2 = 4$, and $m = 2$.

Step 3: Determine the value of n.

We are given ${ }^{n-1} P_3:{ }^n P_4=1: 8$, which can be written as: $\frac{{ }^{n-1} P_3}{{ }^n P_4} = \frac{1}{8}$ Using the permutation formula, we have: $\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{8}$ $\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}} = \frac{1}{8}$ $\frac{(n-1)!}{n!} = \frac{1}{8}$ $\frac{(n-1)!}{n(n-1)!} = \frac{1}{8}$ $\frac{1}{n} = \frac{1}{8}$ Therefore, $n = 8$.

Step 4: Calculate the final expression.

We need to find the value of ${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$. Substituting the values of m and n we found: ${ }^8 P_{2+1}+{ }^{8+1} C_2 = { }^8 P_3 + { }^9 C_2$ Using the permutation and combination formulas: ${ }^8 P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$ ${ }^9 C_2 = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36$ Therefore, ${ }^8 P_3 + { }^9 C_2 = 336 + 36 = 372$

Common Mistakes & Tips

• Remember Pascal's Identity and its application to simplify combination expressions.
• Be careful when simplifying the ratio of permutations; writing out the factorial expansions can help avoid errors.
• Always double-check your arithmetic to avoid simple calculation mistakes.

Summary

We first simplified the inequality involving combinations using Pascal's Identity to find the value of m. Then, we used the given ratio of permutations to find the value of n. Finally, we substituted the values of m and n into the expression ${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$ and calculated the result.

Final Answer

The final answer is \boxed{372}, which corresponds to option (C).