Question

If $^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5$ , then find the value of $r$ ?

Answer 1

Hint: For answering this question we will use the formulae for expansion of combinations that is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and simplify the ratio which we have from the question $^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5$ and derive the value of $r$ . We should remember that the rational numbers can never be assigned in combinations or permutations.

Complete step-by-step answer:
Now considering from the question we have the ratio given as $^{15}{{C}_{r}}{{:}^{15}}{{C}_{r-1}}=11:5$ .
From the basic concept we know the formulae for expansion of combinations that is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . After using this we will have $\dfrac{15!}{r!\left( 15-r \right)!}:\dfrac{15!}{\left( r-1 \right)!\left( 15-\left( r-1 \right) \right)!}=11:5$ .
 From the basic concept we know that the rational numbers can never be assigned in combinations or permutations.
After further simplifying this we will have $\dfrac{1}{r.\left( r-1 \right)!\left( 15-r \right)!}:\dfrac{1}{\left( r-1 \right)!\left( 16-r \right)!}=11:5$ .
By simplifying this we will have $\dfrac{\left( r-1 \right)!\left( 16-r \right)!}{r.\left( r-1 \right)!\left( 15-r \right)!}=11:5$ .
Now we will derive the simplified expression
$\begin{align}
  & \dfrac{\left( 16-r \right).\left( 15-r \right)!}{r.\left( 15-r \right)!}=11:5 \\
 & \Rightarrow \dfrac{\left( 16-r \right)}{r}=11:5 \\
\end{align}$ .
Hence from $\dfrac{\left( 16-r \right)}{r}=11:5$ we can conclude that the value of $r$ is $5$ .

Note: While answering questions of this type we should be sure with the calculations and formulae if by mistake we had used the formulae of permutations which is $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ instead of combinations formulae which is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ then we will have $\dfrac{15!}{\left( 15-r \right)!}:\dfrac{15!}{\left( 15-\left( r-1 \right) \right)!}=11:5$ . Make a note that the rational numbers can never be assigned in combinations or permutations. After simplifying this we will have $\dfrac{1}{\left( 15-r \right)!}:\dfrac{1}{\left( 16-r \right)!}=11:5$ . By further simplifying we will have $\left( 16-r \right)=\dfrac{11}{5}$ . Here we will end up having $r=\dfrac{69}{5}$ which is not possible since rational numbers are not allowed. Hence we should be careful.