- Home
- Standard 11
- Mathematics
6.Permutation and Combination
hard
The number of ordered pairs ( $\mathrm{r}, \mathrm{k}$ ) for which $6 \cdot ^{35} \mathrm{C}_{\mathrm{r}}=\left(\mathrm{k}^{2}-3\right)\cdot{^{36} \mathrm{C}_{\mathrm{r}+1}}$. where $\mathrm{k}$ is an integer, is
A
$3$
B
$2$
C
$4$
D
$6$
(JEE MAIN-2020)
Solution
$6 \times^{35} \mathrm{C}_{\mathrm{r}}=\left(\mathrm{k}^{2}-3\right)^{36} \mathrm{C}_{\mathrm{r}+1}$
$k^{2}-3>0 \Rightarrow k^{2}>3$
$\mathrm{k}^{2}-3=\frac{6 \mathrm{\times}^{35} \mathrm{C}_{\mathrm{r}}}{^{36} \mathrm{C}_{\mathrm{r}+1}}=\frac{\mathrm{r}+1}{6}$
Possible values of $\mathrm{r}$ for integral values of $\mathrm{k},$ are
$r=5,35$
number of ordered pairs are $4$
$(5,2),(5,-2),(35,3),(35,-3)$
Standard 11
Mathematics
