Question
If ${^\text{28}}\text{C}_{2\text{r}}:{^\text{24}}\text{C}_{\text{2r-4}},=225:11,$ Find r.

Answer

We have, $\Rightarrow \frac{\frac{28!}{(2\text{r})!(28-2\text{r}!)}}{\frac{24!}{(2\text{r}-4)!(24-(2\text{r}-4))!}}$ $\Rightarrow \frac{28\times27\times26\times24!(2\text{r}4)!(28-2\text{r})}{(2\text{r})!(28-2\text{r})!24!}=\frac{225}{11}$ $\Rightarrow \frac{28\times27\times25}{2\text{r}\times(2\text{r}-1)\times(2\text{r}-2)(2\text{r}-3)}=2\text{r}(2\text{r-1})(2\text{r}-2)(2\text{r}-3)$ $\Rightarrow 11\times12\times13\times14=2\text{r}(2\text{r}-1)(2\text{r}-2)(2\text{r}-3)$ $\Rightarrow \text{r}=7$

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