If ${ }^{15} \mathrm{C}_4+{ }^{15} \mathrm{C}_5+{ }^{16} \mathrm{C}_6+{ }^{17} \mathrm{C}_7+{ }^{18} \mathrm{C}_8={ }^{19} \mathrm{C}_{\mathrm{r}}$, then the value of r is equal to

Given equation: ${}^{15}C_{4}+{}^{15}C_{5}+{}^{16}C_{6}+{}^{17}C_{7}+{}^{18}C_{8}={}^{19}C_{r}$

Apply Pascal’s Identity ${}^{n}C_{r}+{}^{n}C_{r+1}={}^{n+1}C_{r+1}$ consecutively:

${}^{15}C_{4}+{}^{15}C_{5}={}^{16}C_{5}$

The series becomes ${}^{16}C_{5}+{}^{16}C_{6}+{}^{17}C_{7}+{}^{18}C_{8}$

${}^{16}C_{5}+{}^{16}C_{6}={}^{17}C_{6}$
${}^{17}C_{6}+{}^{17}C_{7}={}^{18}C_{7}$

The expression simplifies to ${}^{18}C_{7}+{}^{18}C_{8}={}^{19}C_{8}$

Thus ${}^{19}C_{8}={}^{19}C_{r}$

This equality holds when $r=8$ or $r=19-8=11$

The solution corresponds to option D.