Question

for some civil cases at least for 9 of 12 jurors must agree on a verdict how many combinations of 9 doors are possible on a 12-person jury?

Answer 1

To solve this, it is necessary to use combinations

`\begin{gathered} 12C9=(12!)/((12-9)!\cdot9!) \\ 12C9=220 \end{gathered}`

There are 220 possible combinations of 9 doors on a 12 person jury