Question

If ${}^n{C}_r=10{,}^n{C}_{r+1}=45$ then, $r$ equals

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

Let ${}^n{C}_r=10\Rightarrow \frac{n!}{r!\left(n-r\right)!}=10$ and
${}^n{C}_{r+1}=45\Rightarrow \frac{n!}{r+1!\left(n-r-1\right)!}=45$
Hence we get,
$\frac{{}^n{C}_{r+1}}{{}^n{C}_r}=\frac{\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}}{\frac{n!}{r!\left(n-r\right)!}}=\frac{n-r}{r+1}=\frac{45}{10}=\frac{9}{2}$
Now we will eliminate the options.
Check that if $r=2$ or $r=4$ then $n$ can not be integer.
For $r=3$, check that ${}^5{C}_3=10$, but ${}^5{C}_4\ne 45$.
Hence the only possible value for $r$ is 1 and $n$ is 10.