Question

Show that$\left(\begin{array}{c}n\\ n-1\end{array}\right)=nand\left(\begin{array}{c}n\\ n\end{array}\right)=1$

Short answer

If j and n are integers with$0\le j\le nthen\left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!}$

Step-by-step solution

  Step 1. Given information 

Here

$\left(\begin{array}{c}n\\ n-1\end{array}\right)and\left(\begin{array}{c}n\\ n\end{array}\right)aregiven$.

we have to show that

$\left(\begin{array}{c}n\\ n-1\end{array}\right)=nand\left(\begin{array}{c}n\\ n\end{array}\right)=1$

Step2.  Showing nn-1=n

We know that

$$\begin{gathered} \left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!},jandnintegers0\le j\le n \\ then\left(\begin{array}{c}n\\ n-1\end{array}\right)=\frac{n!}{(n-n+1)!(n-1)!} \\ =\frac{n!}{1!(n-1)!} \\ =\frac{n(n-1)!}{(n-1)!} \\ =n \end{gathered}$$

Step 3.showing  nn =1

we know that

$$\begin{gathered} \left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!},jandnintegers0\le j\le n \\ then\left(\begin{array}{c}n\\ n\end{array}\right)=\frac{n!}{(n-n)!\left(n\right)!} \\ =\frac{n!}{0!\times n!} \\ =\frac{n!}{n!} \\ =1 \end{gathered}$$$$\begin{gathered} Ifjandnintegers0\le j\le n,then\left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!},jandnintegers0\le j\le n \\ then\left(\begin{array}{c}n\\ n\end{array}\right)=\frac{n!}{(n-n)!\left(n\right)!} \\ =\frac{n!}{0!\left(n\right)!} \\ =\frac{1!}{0!} \\ =1 \end{gathered}$$