Question

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

Answer 1

Note: How many different VOTE COUNTS are possible

Answer 2

11.5 i got but i have no idea what im doing so dont listen to me

Answer 3

13 if divided fairly with a remainder of 1

Answer 4

But...how can we get 13 R 1 vote counts? That's a decimal

Answer 5

it's impossible for 4 to be divided into 50 evenly.

Answer 6

Why do you divide it?

Answer 7

To find the possible amount of votes

Answer 8

50 diffrent vote counts.....and is there any given awnsers

Answer 9

It can vary but I'm dividing it evenly the best I can for four students and 50 votes

Answer 10

its 50 memebers and 4 electives

Answer 11

everyone has to vote

Answer 12

I mean't that ;)

Answer 13

electies*

Answer 14

or whatever the word is ....lol

Answer 15

candidates?

Answer 16

lol derp

Answer 17

so the answer would be and why?

Answer 18

Sorry, I can't think today XD

Answer 19

Lol there are many possible answers. But if it were to be as evenly as possible between the 4 canidates, it would be 13 for the three candidates and 14 for the fourth candidate

Answer 20

But it's asking for the number of vote counts possible.

Answer 21

well if there is 50 members who have to vote... 50 vote counts.

Answer 22

What they mean by vote counts is different ways people can vote for candidate ABC or D.

Answer 23

Wait. But then wouldn't the problem just be asking for the number of ways 50 could be split into 3 groups?

Answer 24

Uhggg...I'm so tired i can't think

Answer 25

well 50 cannot be evenly distributed into 4 groups/ candidates

Answer 26

The question says, "How many different vote counts are possible, if everyone is required to vote?" well 50 if everyone is supposed to vote.

Answer 27

i wanna say (50*49*48*47)*4

Answer 28

I just submited 50 and aaronq's answer into the answer box on my online homework and it says both of them are wrong

Answer 29

lol

Answer 30

sorry :S

Answer 31

This is hard

Answer 32

AUGH!!

Answer 33

I'm going to cry my eyes out in a corner now XD

Answer 34

xD

Answer 35

I'm going to click give up and see what the answer and the solution say.

Answer 36

I would have done that in the first place xD lol jk. sorry :/

Answer 37

Ok... this is going to look weird but:

Let the students receive w,x,y,z votes, respectively. Then we seek all ordered quadruples (w,x,y,z) of nonnegative integers satisfying w+x+y+z = 50. We can think of this as arranging 50 V's (votes) and 3 |'s (dividers), so there are \binom{53}{3}=23426 possible vote tallies.

Answer 38

............... O_O

Answer 39

Oh well! It's an answer! You're right I should have given up earlier :) thanks for trying!

Answer 40

Lol don't give up easily like I do. I suck at math so that's always my answer... ¯\_(ツ)_/¯