1. Write the given expression

We are given

$$\frac{{}^{n+2}C_6}{{}^{n-2}P_2}=11$$

We use:

$${}^{n+2}C_6=\frac{(n+2)!}{6!(n-4)!}$$

and

$${}^{n-2}P_2=\frac{(n-2)!}{(n-4)!}=(n-2)(n-3)$$

2. Substitute into the equation

So,

$$\frac{\dfrac{(n+2)!}{6!(n-4)!}}{(n-2)(n-3)}=11$$

Now expand the factorial in the numerator:

$$(n+2)!=(n+2)(n+1)n(n-1)(n-2)(n-3)(n-4)!$$

Hence,

$${}^{n+2}C_6=\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6!}$$

Therefore,

$$\frac{{}^{n+2}C_6}{{}^{n-2}P_2} =\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6!(n-2)(n-3)}$$ $$=\frac{(n+2)(n+1)n(n-1)}{720}$$

So the equation becomes

$$\frac{(n+2)(n+1)n(n-1)}{720}=11$$

3. Simplify

Multiply both sides by $720$:

$$(n+2)(n+1)n(n-1)=7920$$

Now group terms:

$$(n+2)(n-1)=n^2+n-2$$ $$(n+1)n=n^2+n$$

Let

$$x=n^2+n$$

Then,

$$(n+2)(n+1)n(n-1)=(n^2+n)(n^2+n-2)=x(x-2)$$

So,

$$x(x-2)=7920$$ $$x^2-2x-7920=0$$

4. Solve for $x$

$$x^2-2x-7920=0$$

Discriminant:

$$D=(-2)^2+4\cdot 7920=4+31680=31684=178^2$$

Thus,

$$x=\frac{2\pm 178}{2}$$

So,

$$x=90 \quad \text{or} \quad x=-88$$

Since $x=n^2+n$, we get

$$n^2+n=90$$

or

$$n^2+n=-88$$

The second is impossible for integer/real admissible $n$ here. So,

$$n^2+n-90=0$$

5. Compare with options

We derived:

$$n^2+n-90=0$$

Now check the given options:

• A: $n^2+3n-108=0$
• B: $n^2+5n-84=0$
• C: $n^2+2n-80=0$
• D: $n^2+n-110=0$

None matches $n^2+n-90=0$.

6. Direct verification using the valid value of $n$

From

$$n^2+n-90=0$$

we get

$$(n-9)(n+10)=0$$

So admissible $n=9$.

Check in original expression:

$$\frac{{}^{11}C_6}{{}^{7}P_2}=\frac{462}{42}=11$$

which is correct.

So the actual equation satisfied by $n$ is

$$n^2+n-90=0$$

not any of the listed options.