In this case, we have 6 places to fill from a pool of 22 candidates. There is no difference in the places: they are all commissioners, so if candidate A is selected, the order does not matter, but there is only one candidate, so there is no repetition.
Then, this is a combination (as the order does not matter) with no repetition of 22 elements in 6 places.
We can calculate this as:
![\begin{gathered} C(n,r)=(n!)/(r!(n-r)!) \\ C(22,6)=(22!)/(6!(22-6)!)=(22!)/(6!\cdot14!)=(22\cdot21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15)/(6\cdot5\cdot4\cdot3\cdot2\cdot1)=17907120 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x2i3nevv2k3xnsi501zgs964u5r3dk91eu.png)
Answer: there are 17,907,120 ways this can be done.