Playmakers + coach

In a basketball game, two pivots, two wings, and one point guard play. The coach has three pivots, four wing players, and two playmakers available on the bench. How many different five players can a coach send to the board during a game?

Final Answer:

n =  36

Step-by-step explanation:

$$\begin{gathered} C_2(3)=\left(\genfrac{}{}{0ex}{}{3}{2}\right)=\frac{3!}{2!(3-2)!}=\frac{3}{1}=3 \\ {C}_2(4)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)=\frac{4!}{2!(4-2)!}=\frac{4\cdot 3}{2\cdot 1}=6 \\ {C}_1(2)=\left(\genfrac{}{}{0ex}{}{2}{1}\right)=\frac{2!}{1!(2-1)!}=\frac{2}{1}=2 \\ n=\left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{2}{1}\right)=3\cdot 6\cdot 2=36 \end{gathered}$$

Try another example