6.Permutation and Combination
hard

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 ${ }^{\mathrm{q}+\mathrm{s}} \mathrm{C}_{\mathrm{r}-\mathrm{s}}$ is equal to .... .

A

$136$

B

$1$

C

$16$

D

$13$

(JEE MAIN-2021)

Solution

${ }^{1} \mathrm{P}_{1}+2 \cdot{ }^{2} \mathrm{P}_{2}+3 \cdot{ }^{3} \mathrm{P}_{3}+\ldots+15 \cdot{ }^{15} \mathrm{P}_{15}$

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

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

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

$=16 !-1$

$={ }^{16} \mathrm{P}_{16}-1$

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

${ }^{\mathrm{q}+\mathrm{s}} \mathrm{C}_{\mathrm{r}-\mathrm{s}}={ }^{17} \mathrm{C}_{15}=136$

Standard 11
Mathematics

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