Question

If ${}^{15}{C}_{3r}{:}^{15}{C}_{r+1}=11:3$ find $r$.

Answer 1

$\frac{{}^{15}{C}_{3r}}{{}^{15}{C}_{r+1}}=\frac{11}{3}$ ${}^n{C}_r=\frac{n!}{r!(n-r)!}$

$\frac{\frac{15!}{\left(3r\right)!(15-3r)!}}{\frac{15!}{(r+1)!(14-r)!}}=\frac{11}{3}$

when $r=1$

$LHS=\frac{\frac{15!}{3!\times 12!}}{\frac{15!}{4!\times 11!}}=\frac{4!\times 11!}{3!\times 12!}=\frac{4\times 3!\times 11!}{3!\times 12\times 11!}=\frac{1}{3}$

$\ne RHS$

when $r=2$

$LHS=\frac{\frac{15!}{6!\times 9!}}{\frac{15!}{3!\times 12!}}=\frac{3!\times 12\times 11\times 10\times 9!}{6\times 5\times 4\times 3!\times 9!}=11\ne RHS$

when $r=3$

$LHS=\frac{\frac{15!}{9!\times 6!}}{\frac{15!}{4!\times 11!}}=\frac{11\times 4!\times 10\times 9!}{6\times 5\times 4!\times 9!}=\frac{11}{3}=RHS$

Thus $r=3$