+0

+1
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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?

Jul 12, 2020

#1
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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.

Jul 12, 2020
#2
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Sorry about that, I wasn't online :(

I'll take a look at the others

Jul 12, 2020
#3
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Jul 12, 2020
#4
+111566
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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.

Jul 12, 2020
#5
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I'll use the second method....

Would it be 7C5/16C5?

Possibilities/Total?

Jul 12, 2020
#6
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Have you read the question or tried to think it through?

It says HOW MANY

Why would that be a fraction?

Melody  Jul 12, 2020
#7
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The total number of ways would be binomial (16 5)

Jul 12, 2020
edited by CalTheGreat  Jul 12, 2020
#9
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Yes but that is not the question Cal  AND  I want Helpbot to answer and not just be totally spoonfed.

Melody  Jul 12, 2020
#10
+111566
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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.

Jul 12, 2020
#12
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Yeah, you're right, Melody.

When HB sees this, I hope he/she'll be able to figure it out!

CalTheGreat  Jul 12, 2020
#13
+461
+1

I think I got it.

I subtracted the total possibilities from the possibilities and the number of ways to choose officers without any of the past to get 4242.

Thank you for all of your help, and sorry if I was a pain :(

Jul 12, 2020