Question

If n and r are integers such that $1\le r\le n$, then $n.C\left( n-1,r-1 \right)=$
A. $C\left( n,r \right)$
B. $n.C\left( n,r \right)$
C. $rC\left( n,r \right)$
D. $\left( n-1 \right).C\left( n,r \right)$

Answer 1

Hint: We will start by using the fact that $C\left( n,r \right)$ is another way of writing ${}^{n}{{C}_{r}}$. Then we will use the fact that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$ to find the value of $n.C\left( n-1,r-1 \right)$. Then we will multiply it by r in both numerator and denominator and find the answer.

Complete step-by-step answer:
Now, we have to find the value of $n.C\left( n-1,r-1 \right)$.
Now, we need to understand that $C\left( n,r \right)$ is another way writing $C\left( n,r \right)={}^{n}{{C}_{r}}$. So, we have $n.C\left( n-1,r-1 \right)=n.{}^{n-1}{{C}_{r-1}}$.
Now, we know that the fact ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$.
So, we have $n.{}^{n-1}{{C}_{r-1}}$ as,
$\begin{align}
  & \Rightarrow \dfrac{n.\left( n-1 \right)!}{\left( n-1-r+1 \right)!\left( r-1 \right)!} \\
 & \Rightarrow \dfrac{n.\left( n-1 \right)!}{\left( n-r \right)!\times \left( r-1 \right)!} \\
\end{align}$
Now, we know that $n\times \left( n-1 \right)!=n!$. So, we have,
$\Rightarrow \dfrac{n!}{\left( n-r \right)!\times \left( r-1 \right)!}$
Now, we multiply both numerator and denominator by r. So, we have,
$\begin{align}
  & \Rightarrow \dfrac{r.n!}{\left( n-r \right)!\times \left( r-1 \right)!\times r} \\
 & \Rightarrow \dfrac{r.n!}{\left( n-r \right)!\times r!} \\
\end{align}$
Now, we know that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$. So, we have,
$\begin{align}
  & r.\dfrac{n!}{\left( n-r \right)!\times r!}=r.{}^{n}{{C}_{r}} \\
 & =r.C\left( n,r \right) \\
\end{align}$
Hence, the correct option is (C).

Note: It is important to note that we have used a fact that ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$. Also, it is important to note that we have multiplied $\dfrac{n!}{\left( n-r \right)!\times \left( r-1 \right)!}$ by r in numerator and denominator because by doing this we will have $r!$ in denominator and hence we will have the expanded form of ${}^{n}{{C}_{r}}$. So, we must know that whenever we get this type of question, we can multiply with the right term to get the result asked in the question.