Chapter 12, Problem 75RE

To determine

Evaluate the expression ${\displaystyle \sum _{k=0}^5\left(\begin{array}{l}5\\ k\end{array}\right)}$ .

Answer to Problem 75RE

The evaluate value of the expression is $32$

Explanation of Solution

Given information:

It is given in the question that, the expression is ${\displaystyle \sum _{k=0}^5\left(\begin{array}{l}5\\ k\end{array}\right)}$ .

Formula used:

  $\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Calculation:

  ${\displaystyle \sum _{k=0}^5\left(\begin{array}{l}5\\ k\end{array}\right)}$

By using the formula $\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

  $\begin{array}{l}{\displaystyle \sum _{k=0}^5\left(\begin{array}{l}5\\ k\end{array}\right)}\\ \frac{5!}{0!\left(5-0\right)!}+\frac{5!}{1!\left(5-1\right)!}+\frac{5!}{2!\left(5-2\right)!}+\frac{5!}{3!\left(5-3\right)!}+\frac{5!}{4!\left(5-4\right)!}+\frac{5!}{5!\left(5-5\right)!}\\ \frac{5!}{0!*\left(5\right)!}+\frac{5!}{1!*\left(4\right)!}+\frac{5!}{2!*\left(3\right)!}+\frac{5!}{3!*\left(2\right)!}+\frac{5!}{4!*\left(1\right)!}+\frac{5!}{5!*\left(0\right)!}\\ \frac{5!}{1*\left(5\right)!}+\frac{5*4!}{1*4!}+\frac{5*4*3!}{2*\left(3\right)!}+\frac{5*4*3!}{3!*2*1}+\frac{5*4!}{4!*\left(1\right)!}+\frac{5!}{5!*1}\\ 1+5+5*2+5*2+5+1\\ 32\end{array}$