## College Algebra 7th Edition

$\left(\begin{array}{l} 5\\ 0 \end{array}\right)+\left(\begin{array}{l} 5\\ 1 \end{array}\right)+\left(\begin{array}{l} 5\\ 2 \end{array}\right)+\left(\begin{array}{l} 5\\ 3 \end{array}\right)+\left(\begin{array}{l} 5\\ 4 \end{array}\right)+\left(\begin{array}{l} 5\\ 5 \end{array}\right)=32$
We evaluate: $\left(\begin{array}{l} 5\\ 0 \end{array}\right)+\left(\begin{array}{l} 5\\ 1 \end{array}\right)+\left(\begin{array}{l} 5\\ 2 \end{array}\right)+\left(\begin{array}{l} 5\\ 3 \end{array}\right)+\left(\begin{array}{l} 5\\ 4 \end{array}\right)+\left(\begin{array}{l} 5\\ 5 \end{array}\right)$ We notice that this represents an expanded 5th power binomial. Instead of evaluating each term, we simply reverse the expansion into the original binomial: $=(1+1)^{5}=2^{5}=32$