At the end of the year, the Math Club decided to hold an election for which 5 equal officer positions were available. However, 16 candidates were nominated, of whom 7 were past officers. Of all possible elections of the officers, how many will have at least 1 of the past officers?
Method1:
9 new members and 7 that have been members before. 16 altogether
5 positions.
they can all come from the past members that is 7C5 ways
You can have one from the new and 4 from the old. That is 9C1*7C4
etc
OR
method 2
there are 16C5 possible selections
9C5 of those would have no old member
the rest would have at least one.
The second method is easier.
You work it out from here and then show us your working/logic.