If $\frac{^{15} C_{r}}{^{15} C_{r - 1}} = \frac{11}{5}$ find $r .$

Open in App

Solution

Given that,

$\frac{^{15} C_{r}}{^{15} C_{r - 1}} = \frac{11}{5}$ ......(1)

Since, $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$

Substituting $n = 15$ in equation(1), we have

$\frac{15 - r + 1}{r} = \frac{11}{5}$

$75 - 5 r + 5 = 11 r$

$16 r = 80$

$r = 5$

Hence, the value of $r$ is $5$ .

Suggest Corrections

Similar questions

^{15} C_{r} :^{15} C_{r - 1} = 11 : 5

If $^{15} C_{r} :^{15} C_{r - 1} = 11 : 5$ , find r.

^{15} C_{3 r} :^{15} C_{r + 1} = 11 : 3

r

^{15} C_{3 r} :^{15} C_{r + 1} = 11 : 3

r

(^{15} C_{r} +^{15} C_{r - 1}) (^{15} C_{15 - r} +^{15} C_{16 - r}) = (^{16} C_{13})^{2}

r

Join BYJU'S Learning Program