Question

Given that the value of ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r}}$ then the value of x is equal to
\[\begin{align}
  & \text{A}.\text{ r} \\
 & \text{B}.\text{ r}+\text{1} \\
 & \text{C}.\text{ r}-\text{1} \\
 & \text{D}.\text{ 2r} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will use a property or formula of ${}^{n}{{C}_{r}}$ which is stated as below:
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\]
We will substitute this value of r as r+1 in above and then try to compare it with ... to get the result.

Complete step by step answer:
Given that, \[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n}{{C}_{x}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
We will first of all define what is ${}^{n}{{C}_{r}}$
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
We also have a property of ${}^{n}{{C}_{r}}$ relation defined as below:
${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$
We will use this equation (ii) to find value of x in equation (i).
Consider equation (ii).
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}\]
Substituting r=r+1 in above we get:
\[\begin{align}
  & \Rightarrow {}^{n}{{C}_{r+1}}+{}^{n}{{C}_{r+1-1}}={}^{n+1}{{C}_{r+1}} \\
 & \Rightarrow {}^{n}{{C}_{r+1}}+{}^{n}{{C}_{r}}={}^{n+1}{{C}_{r+1}} \\
\end{align}\]
Which is nothing but
\[{}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Now, comparing equation (i) and (ii) we see that, LHS of both equation are same
\[\Rightarrow {}^{n}{{C}_{r+1}}={}^{n}{{C}_{x}}\]
Comparing the subscript of above both sides, we get:
\[\begin{align}
  & r+1=x \\
 & \Rightarrow x=r+1 \\
\end{align}\]
Hence, the value of x is r+1. So, option B is the correct answer.

Note:
A possible mistake in this question can be expanding equation (i) by using ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ this may lead to confusion at the end because we are not aware of the value of x and hence expanding ${}^{n}{{C}_{x}}$ would be tricky and involves a lot of calculation mistake. So, the above-stated formula should be followed.