Solveeit Logo
Login

Question

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...

If n and r are integers such that 1≤r≤n1\le r\le n, then n.C(n−1,r−1)=n.C\left( n-1,r-1 \right)=
A. C(n,r)C\left( n,r \right)
B. n.C(n,r)n.C\left( n,r \right)
C. rC(n,r)rC\left( n,r \right)
D. (n−1).C(n,r)\left( n-1 \right).C\left( n,r \right)

Explanation

Solution

Hint: We will start by using the fact that C(n,r)C\left( n,r \right) is another way of writing nCr{}^{n}{{C}_{r}}. Then we will use the fact that nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!} to find the value of n.C(n−1,r−1)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(n−1,r−1)n.C\left( n-1,r-1 \right).
Now, we need to understand that C(n,r)C\left( n,r \right) is another way writing C(n,r)=nCrC\left( n,r \right)={}^{n}{{C}_{r}}. So, we have n.C(n−1,r−1)=n.n−1Cr−1n.C\left( n-1,r-1 \right)=n.{}^{n-1}{{C}_{r-1}}.
Now, we know that the fact nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}.
So, we have n.n−1Cr−1n.{}^{n-1}{{C}_{r-1}} as,
⇒n.(n−1)!(n−1−r+1)!(r−1)! ⇒n.(n−1)!(n−r)!×(r−1)! \begin{aligned} & \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{aligned}
Now, we know that n×(n−1)!=n!n\times \left( n-1 \right)!=n!. So, we have,
⇒n!(n−r)!×(r−1)!\Rightarrow \dfrac{n!}{\left( n-r \right)!\times \left( r-1 \right)!}
Now, we multiply both numerator and denominator by r. So, we have,
⇒r.n!(n−r)!×(r−1)!×r ⇒r.n!(n−r)!×r! \begin{aligned} & \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{aligned}
Now, we know that nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}. So, we have,
r.n!(n−r)!×r!=r.nCr =r.C(n,r) \begin{aligned} & r.\dfrac{n!}{\left( n-r \right)!\times r!}=r.{}^{n}{{C}_{r}} \\\ & =r.C\left( n,r \right) \\\ \end{aligned}
Hence, the correct option is (C).

Note: It is important to note that we have used a fact that nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}. Also, it is important to note that we have multiplied n!(n−r)!×(r−1)!\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!r! in denominator and hence we will have the expanded form of nCr{}^{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.