If ${ }^{n+4} C_{n+1}-{ }^{n+3} C_n=15(n+2)$, then $n=$

Applying the symmetry property ${}^N C_R = {}^N C_{N-R}$, we rewrite:

${}^{n+4} C_{n+1} = {}^{n+4} C_3$
${}^{n+3} C_n = {}^{n+3} C_3$

The original equation becomes:

${}^{n+4} C_3 - {}^{n+3} C_3 = 15(n+2)$

Using the identity ${}^N C_R - {}^{N-1} C_R = {}^{N-1} C_{R-1}$ with $N = n+4$ and $R = 3$:

${}^{n+4} C_3 - {}^{n+3} C_3 = {}^{n+3} C_2$

Thus, the equation simplifies to:

${}^{n+3} C_2 = 15(n+2)$

Expanding the combination:

${}^{n+3} C_2 = \frac{(n+3)(n+2)}{2}$

Substituting back:

$\frac{(n+3)(n+2)}{2} = 15(n+2)$

Since $n \ge 0$ for valid combinations, we can divide both sides by $(n+2)$:

$\frac{n+3}{2} = 15$

Solving yields $n = 27$.

Final answer: $\boxed{D}$