1. Find $m$ from the combination inequality

Given

$${}^6C_m+2\left({}^6C_{m+1}\right)+{}^6C_{m+2} > {}^8C_3$$

First compute:

$${}^8C_3 = 56$$

Now test integer values of $m$ for which the binomial coefficients are defined.

Using row 6 of Pascal's triangle:

$${}^6C_0,{}^6C_1,{}^6C_2,{}^6C_3,{}^6C_4,{}^6C_5,{}^6C_6 = 1,6,15,20,15,6,1$$

Let

$$S(m)={}^6C_m+2{}^6C_{m+1}+{}^6C_{m+2}$$

Now check values:

• For $m=0$:

$$S(0)=1+2(6)+15=28$$

• For $m=1$:

$$S(1)=6+2(15)+20=56$$

• For $m=2$:

$$S(2)=15+2(20)+15=70$$

Since the inequality is strict, $S(m)>56$ gives

$$m=2$$

(as the required value fitting the condition).

2. Find $n$ from the permutation ratio

Given

$${}^{n-1}P_3 : {}^nP_4 = 1:8$$

So,

$$\frac{{}^{n-1}P_3}{{}^nP_4}=\frac18$$

Using

$${}^{n-1}P_3=(n-1)(n-2)(n-3)$$

and

$${}^nP_4=n(n-1)(n-2)(n-3)$$

Therefore,

$$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac1n=\frac18$$

Hence,

$$n=8$$

3. Compute the required expression

We need

$${}^nP_{m+1}+{}^{n+1}C_m$$

With $n=8$ and $m=2$:

$${}^8P_3+{}^9C_2$$

Now,

$${}^8P_3=8\cdot 7\cdot 6=336$$

and

$${}^9C_2=\frac{9\cdot 8}{2}=36$$

So the sum is

$$336+36=372$$

4. Final answer

$$\boxed{372}$$

This corresponds to Option C.