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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?

 Apr 30, 2015

Best Answer 

 #5
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+9

Melody, I agree with your answers –both of them.

 

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.

 May 3, 2015
 #1
avatar+110208 
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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.

 

When you find the answer can you tell us please ?

 May 3, 2015
 #2
avatar+1807 
0

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 :(

 May 3, 2015
 #3
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+5

Hold on, I'll work on it for you.   . . . . 

 May 3, 2015
 #4
avatar+110208 
+7

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://web.eecs.utk.edu/~booth/311-04/homeworks/hw2sol.pdf    Question 4

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

 May 3, 2015
 #5
avatar
+9
Best Answer

Melody, I agree with your answers –both of them.

 

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
 #6
avatar+110208 
+4

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 

 May 3, 2015
 #7
avatar+1807 
+5

Yes!!! This is correct!! Thank you guys sooo sooo much!!

 May 3, 2015
 #8
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0

yes it is right

 May 2, 2016
 #9
avatar+6 
-1

Yep. Mellie taking courses on AoPS. Confirmed.

 Aug 11, 2016

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