College Algebra, Chapter 9, 9.6, Section 9.6, Problem 24

Evaluate the expression
$
\left(
\begin{array}{c}
5\\
0
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
1
\end{array}
\right)
+
\left(
\begin{array}{c}
5\\
2
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
3
\end{array}
\right)
+
\left(
\begin{array}{c}
5\\
4
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
5
\end{array}
\right)
$

Recall that the binomial coefficient is denoted by $\displaystyle \left( \frac{n}{r} \right)$ and is defined by
Notice that these are precisely the entries in the fifth row of Pascal's Triangle.

$
\begin{equation}
\begin{aligned}
\left(
\begin{array}{c}
5\\
0
\end{array}
\right)
&=
\frac{5!}{0!(5-0)!} = 1\\
\\

\left(
\begin{array}{c}
5\\
1
\end{array}
\right)
&=
\frac{5!}{1!(5-1)!} = 5\\
\\

\left(
\begin{array}{c}
5\\
2
\end{array}
\right)
&=
\frac{5!}{2!(5-2)!} = 10\\
\\

\left(
\begin{array}{c}
5\\
3
\end{array}
\right)
&=
\frac{5!}{3!(5-3)!} = 10\\
\\

\left(
\begin{array}{c}
5\\
4
\end{array}
\right)
&=
\frac{5!}{4!(5-4)!} = 5 \\
\\

\left(
\begin{array}{c}
5\\
5
\end{array}
\right)
&=
\frac{5!}{5!(5-5)!} = 1

\end{aligned}
\end{equation}
$

Thus,

$
\displaystyle
\left(
\begin{array}{c}
5\\
0
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
1
\end{array}
\right)
+
\left(
\begin{array}{c}
5\\
2
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
3
\end{array}
\right)
+
\left(
\begin{array}{c}
5\\
4
\end{array}
\right)
-
\left(
\begin{array}{c}
5\\
5
\end{array}
\right)
= 1 - 5 + 10 - 10 + 5 - 1 = 0
$