Question:
If ${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}={ }^{9} \mathrm{P}-\mathrm{s}, 0 \leq \mathrm{s} \leq 1$
then ${ }^{4+5} C_{x-s}$ is equal to_________
Solution:
${ }^{1} P_{1}+2 \cdot{ }^{2} P_{2}+3 \cdot{ }^{3} P_{3}+\ldots+15 \cdot{ }^{15} P_{15}$
$=1 !+2.2 !+3.3 !+\ldots .15 \times 15 !$
$=\sum_{r=1}^{15}(r+1-1) r !$
$=\sum_{r=1}^{15}(r+1) !-(r) !$
$=16 !-1$
$={ }^{16} P_{16}-1$
$\Rightarrow \mathrm{q}=\mathrm{r}=16, \mathrm{~s}=1$
${ }^{4+5} \mathrm{C}_{t-6}={ }^{17} \mathrm{C}_{15}=136$