Numbered

In the past, passengers on public transport validated single-use tickets that had 9 numbered boxes, a certain number of which were punched with a validator.
A) In how many different ways could a ticket be validated if 3 boxes were punched?
B) In how many ways could a ticket be validated if 6 boxes were punched?

Final Answer:

n₁ =  84
n₂ =  84

Step-by-step explanation:

$$\begin{gathered} C_3(9)=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=\frac{9!}{3!(9-3)!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=84 \\ {n}_1=\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84 \end{gathered}$$

$$\begin{gathered} C_6(9)=\left(\genfrac{}{}{0ex}{}{9}{6}\right)=\frac{9!}{6!(9-6)!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1}=84 \\ {n}_2=\left(\genfrac{}{}{0ex}{}{9}{6}\right)=84 \end{gathered}$$

Try another example