+479 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?
+118608 I was about to answer, but I looked at your record first.
You give your self points but you never have given an answerer any point.
You have not responded to your answerers in any way, other than "That is incorrect"
EP gave you a full answer that Cal responded to. You didn't though.
so I am not inspired to help you.
+118608 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.
+479 Thanks... got a notification that someone answered!
I'll use the second method....
Would it be 7C5/16C5?
Possibilities/Total?
+118608 Cal please give Helpbot time and incentive to work it out for him/her self.
It is good that you are thinking, and your hint was good, I just do not want helpbot to be given that much help yet.
+2094 Yeah, you're right, Melody.
Sorry about that.
When HB sees this, I hope he/she'll be able to figure it out!
