Question:
If ${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}={ }^{\mathrm{q}} \mathrm{P}_{\mathrm{r}}-\mathrm{s}, 0 \leq \mathrm{s} \leq 1$ then ${ }^{q+s} \mathrm{C}_{r-s}$ is equal to _______.
Solution:
${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}$
$=1 !+2.2 !+3.3 !+\ldots .15 \times 15 !$
$=\sum_{r=1}^{15}(r+1-1 r !)$
$=\sum_{r=1}^{15}(r+1) !-(r)$
$=16 !-1$
$={ }^{16} \mathrm{P}_{16}-1$
$\Rightarrow \mathrm{q}=\mathrm{r}=16, \mathrm{~s}=1$
${ }^{\mathrm{q}+\mathrm{s}} \mathrm{C}_{\mathrm{r}-\mathrm{s}}={ }^{17} \mathrm{C}_{15}=136$