Question

If ${}^{15}{C}_{3r}{:}^{15}{C}_{r+1}=11:3$ then the value of $r$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

${}^{15}{C}_{3r}{:}^{15}{C}_{r+1}=13:3$.

$\frac{15!}{(15-3r)!\left(3r\right)!}\times \frac{(15-r-1)!(r+1)!}{15!}=\frac{13}{3}$

$\frac{(15-r-1)!(r+1)!}{(15-3r)!\left(3r\right)!}=\frac{11}{3}$

$0\le 3r\le 15$

$0\le r\le 5$

$\therefore$ By hit and trial method

let r=1

$\frac{13!2!}{12!3!}=\frac{13}{3}$

Answer=1