Evaluate using Pascal's triangle. $\left( \begin{array} { l }
6 \
5
\end{array} \right)$
A) $\left( \begin{array} { l }
6 \
5
\end{array} \right) = 720$
B) $\left( \begin{array} { l }
6 \
5
\end{array} \right) = 21$
C) $\left( \begin{array} { l }
6 \
5
\end{array} \right) = 56$
D) $\left( \begin{array} { l }
6 \
5
\end{array} \right) = 1$
E) $\left( \begin{array} { l }
6 \
5
\end{array} \right) = 6$

Question

Evaluate using Pascal's triangle.​ $\left( \begin{array} { l } 
6 \
5
\end{array} \right)$ ​
A) $\left( \begin{array} { l } 
6 \
5
\end{array} \right) = 720$
B) $\left( \begin{array} { l } 
6 \
5
\end{array} \right) = 21$
C) $\left( \begin{array} { l } 
6 \
5
\end{array} \right) = 56$
D) $\left( \begin{array} { l } 
6 \
5
\end{array} \right) = 1$
E) $\left( \begin{array} { l } 
6 \
5
\end{array} \right) = 6$

Quizplus · Accepted Answer

$\left( \begin{array} { l } 6 \5\end{array} \right)$ is equivalent to $C(6, 5)$, which calculates the number of combinations of 6 items taken 5 at a time. According to Pascal's triangle or the combination formula, $C(n, k) = \frac{n!}{k!(n-k)!}$, this evaluates to 6.

Evaluate Using Pascal's Triangle (65)\left( \begin{array} { l } 6 \\5\end{array} \right)(65​)

Evaluate Using Pascal's Triangle $\left( \begin{array} { l } 6 \\5\end{array} \right)$