Question

If ${}^{10}C_x={}^{10}C_{x+4}$, then prove that $x=3$.

Answer 1

Step 1: Interpret the given data.

Given, ${}^{10}\mathrm{C}_x={}^{10}\mathrm{C}_{x+4}$.

In combinations,

${}^{n}C_r={}^{n}C_{n-r}$

Here,

$\begin{array}{cc}n=10& \\ r=x& \dots \left(\mathrm{i}\right)\\ n-r=x+4& \dots \left(\mathrm{ii}\right)\end{array}$

Step 2: Add equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$.

Adding the equations,

$\begin{array}{rcl}& & \begin{array}{ccc}\Rightarrow & r+n-r=x+x+4& \\ \Rightarrow & n=2x+4& \\ \Rightarrow & 10=2x+4& \left[\because n=10\right]\\ \Rightarrow & 2x=6& \\ \Rightarrow & x=3& \end{array}\end{array}$

Hence, it is proved that the value of $x=3$, if ${}^{10}\mathrm{C}_x={}^{10}\mathrm{C}_{x+4}$.