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# 1. 4 students are running for club president in a club with 50 members. How many different vote counts are possible, if all 50 members

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1.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if all 50 members are required to vote?

2.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

Mellie  Apr 30, 2015

#5
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I’m Nauseated BTW -- The Troll, not my physical condition.

The network connection keeps dropping out, it connects for less than a minute; I can’t login. I’ve tried to post just this over 15 times.

Guest May 3, 2015
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#1
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I think that this is very like this question

http://web2.0calc.com/questions/in-how-many-ways-can-we-distribute-13-pieces-of-identical-candy-to-5-kids-if-the-two-youngest-kids-are-twins-and-insist-on-receiving-an-equ

You told me my last one was wrong so i don't know.  sorry.

Melody  May 3, 2015
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Yes, but only when I get the whole question wrong three times, will I be able to get it OR if I get the question right. I don't know what to do, because if I get it wrong, my report for the week will be very bad :(

Mellie  May 3, 2015
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Hold on, I'll work on it for you.   . . . .

Guest May 3, 2015
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I think

the number of ways that n identical votes can be distibuted to k people (assuming that each person gets at least one vote =  $$\binom{n+k-1}{n}$$

1.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if all 50 members are required to vote?

I am going to assume that everyone gets at least 1 vote.

(50+4-1)C50 = 53C50

$${\left({\frac{{\mathtt{53}}{!}}{{\mathtt{50}}{!}{\mathtt{\,\times\,}}({\mathtt{53}}{\mathtt{\,-\,}}{\mathtt{50}}){!}}}\right)} = {\mathtt{23\,426}}$$

2.  4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote?

I think this assumes that they all get at least 1 vote and at least 1 does not vote at all

(50+5-1)C50 = 54C50

$${\left({\frac{{\mathtt{54}}{!}}{{\mathtt{50}}{!}{\mathtt{\,\times\,}}({\mathtt{54}}{\mathtt{\,-\,}}{\mathtt{50}}){!}}}\right)} = {\mathtt{316\,251}}$$

If some people get no votes it would be higher - does it need to be worked out that way?

Ref:

http://euclid.ucc.ie/pages/MATHENR/Exercises/CombinatoricsRepetitionsConditions.pdf

Melody  May 3, 2015
#5
+9

I’m Nauseated BTW -- The Troll, not my physical condition.

The network connection keeps dropping out, it connects for less than a minute; I can’t login. I’ve tried to post just this over 15 times.

Guest May 3, 2015
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Thanks Nauseated,

I am actually wondering if my first answer - which I nutted out without a formula was also correct.

It started with a different premise - that a child could get no lollies, that may be the only reason it was 'wrong'.

http://web2.0calc.com/questions/in-how-many-ways-can-we-distribute-13-pieces-of-identical-candy-to-5-kids-if-the-two-youngest-kids-are-twins-and-insist-on-receiving-an-equ#r1

If it was correct then I would have reason to be very pleased with myself.

I guess I will never know

Melody  May 3, 2015
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Yes!!! This is correct!! Thank you guys sooo sooo much!!

Mellie  May 3, 2015
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yes it is right

Guest May 2, 2016
#9
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Yep. Mellie taking courses on AoPS. Confirmed.

Detective  Aug 11, 2016

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