Solve this

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:

${ }^{1} P_{1}+2 \cdot{ }^{2} P_{2}+3 \cdot{ }^{3} P_{3}+\ldots+15 \cdot{ }^{15} P_{15}$

$=1 !+2.2 !+3.3 !+\ldots .15 \times 15 !$

$=\sum_{r=1}^{15}(r+1-1) r !$

$=\sum_{r=1}^{15}(r+1) !-(r) !$

$=16 !-1$

$={ }^{16} P_{16}-1$

$\Rightarrow \mathrm{q}=\mathrm{r}=16, \mathrm{~s}=1$

${ }^{4+5} \mathrm{C}_{t-6}={ }^{17} \mathrm{C}_{15}=136$

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