Question

If ${{}^{28}C}_{2r}+{{}^{24}C}_{2r-4}=225:11$, then $r=$

• A. $24$
• B. $14$
• C. $7$
• D. $noneofthese$

Answer 1

The correct option is C $7$

$\Rightarrow \frac{\frac{\left(2r\right)!}{\left(2r\right)!(28-2r)!}}{\frac{\left(24\right)!}{(28-2r)!(2r-4)!}}=\frac{225}{11}$

$\Rightarrow \frac{28\times 27\times 26\times 25\times (2r-4)!}{2r(2r-1)(2r-2)(2r-3)(2r-4)!}=\frac{225}{11}$

$\therefore r=7$.