Combination Exercise Level 1 asked in the brilliant.org

Exercise:

Solve , $\left( \begin{matrix}
   n  \\
   1234  \\
\end{matrix} \right)=\left( \begin{matrix}
   n  \\
   4321  \\
\end{matrix} \right)$

Solution: we know that $\left( \begin{matrix}
   n  \\
   r  \\
\end{matrix} \right)=\frac{n!}{\left( n-r \right)!r!}$  &  $\left( \begin{matrix}
   n  \\
   a  \\
\end{matrix} \right)=\left( \begin{matrix}
   n  \\
   n-a  \\
\end{matrix} \right)$

So $a=1234\,\,\And \,\,n-a=4321$ $\Rightarrow n-1234=4321\Rightarrow n=4321+1234=5555$

Or we can solve it using the definition of combination as follows :

So $\left( \begin{matrix}
   n  \\
   1234  \\
\end{matrix} \right)=\frac{n!}{\left( n-1234 \right)!1234!}\,\,\,\And \,\,\left( \begin{matrix}
   n  \\
   4321  \\
\end{matrix} \right)=\frac{n!}{\left( n-4321 \right)!4321!}$

Hence $\frac{n!}{\left( n-1234 \right)!1234!}=\frac{n!}{\left( n-4321 \right)!4321!}$

$\Leftrightarrow n!\left( n-4321 \right)!4321!=n!\left( n-1234 \right)!1234!$

$\Rightarrow \frac{\left( n-4321 \right)!4321!}{\left( n-1234 \right)!1234!}=1$

//$\frac{\left( n-a \right)!a!}{\left( n-b \right)!b!}=1$ Put $b=n-a$ $\Rightarrow \frac{\left( n-a \right)!a!}{\left( n-\left( n-a \right) \right)!\left( n-a \right)!}=\frac{\left( n-a \right)!a!}{a!\left( n-a \right)!}=1$

So it’s clear that $n=a+b=5555$