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1. 3 students are running for a class president in a class of 70 students. How many different vote counts are possible if some student(s) do not vote.

2. In how many ways can we distribute 7 pieces of identical taffy and 8 pieces of identical licorice to 5 kids such that each kid recieves exactly 3 pieces of candy.

 

1. I am completely stuck. I have thought of using the "hockey stick identity", but I do not fully understand how. Thanks!

2. I tried listing out all of the possible cases, but there were way too much.

 Mar 4, 2023
edited by iambog  Mar 4, 2023
 #1
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1.  

Let's assume that the 3 candidates are named A, B, and C.

If all 70 students vote, there are a total of 70 choices for the first vote, 69 choices for the second vote, and 68 choices for the third vote, since each student can only vote once and cannot vote for themselves. Therefore, the total number of possible vote counts is:

70 x 69 x 68 = 328,920

Now, if some students do not vote, the total number of possible vote counts will decrease. Let's consider some scenarios:

If one student does not vote: There are 70 choices for who does not vote. For the remaining 69 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when one student does not vote is:

70 x 328,920 = 23,064,400

If two students do not vote: There are (70 choose 2) = 2,415 ways to choose two students out of 70 who do not vote. For the remaining 68 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when two students do not vote is:

2,415 x 328,920 = 793,556,800

If three students do not vote: There is only one way to choose three students out of 70 who do not vote. For the remaining 67 students, there are 328,920 possible vote counts. Therefore, the total number of possible vote counts when three students do not vote is:

1 x 328,920 = 328,920

In summary, the total number of possible vote counts when some students do not vote is:

23,064,400 + 793,556,800 + 328,920 = 817,950,120

 

2. 

We can use a stars and bars approach to solve this problem. We need to distribute 7 pieces of taffy and 8 pieces of licorice to 5 kids such that each kid receives exactly 3 pieces of candy.

Let's represent the distribution using 5 variables: x1, x2, x3, x4, x5, where xi represents the number of pieces of candy given to the ith kid. Since each kid must receive 3 pieces of candy, we can subtract 3 from each xi to simplify the problem.

So, we need to distribute (7-3x5) = 2 pieces of taffy and (8-3x5) = -7 pieces of licorice among 5 kids, without any restriction on the number of pieces each kid gets. Notice that we have negative pieces of licorice, but this simply means that some kids will get no licorice at all.

Using the stars and bars formula, the number of ways to distribute 2 pieces of taffy and -7 pieces of licorice among 5 kids is:

C(2+5-1, 5-1) * C(-7+5-1, 5-1) = C(6, 4) * C(-3, 4)

where C(n, k) represents the number of ways to choose k items out of n.

Since C(-3, 4) is not defined, we need to use a different formula for negative values. The formula is:

C(-n+k-1, k-1) = C(n+k-1, k-1)

Using this formula, we get:

C(6, 4) * C(3+4-1, 4-1) = C(6, 4) * C(6, 3)

= 15 * 20

= 300

Therefore, there are 300 ways to distribute 7 pieces of identical taffy and 8 pieces of identical licorice to 5 kids such that each kid receives exactly 3 pieces of candy.

 Mar 4, 2023
 #2
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For the first question you must have misunderstood. The kids who do not vote are not distinguishable and the amount of kids who do not vote can be any number from 0-70. So every could vote or no one could vote. We are basically finding the amount of ways we can sum to every number 0-70 using three integers and the number of DIFFERENT ways to arrange the vote counts. For example (3,3,3) there is only 1 arrangement, but (3,2,3) has 3 arrangements.

Guest Mar 5, 2023
 #3
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Also for the second, it is the number of ways you can split the identical taffy and licorice to 5 different kids. So we need to solve for the number of ways each kid receives 3 candies 

Guest Mar 5, 2023
 #4
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Please just post one question at a time. 

 

1. 3 students are running for a class president in a class of 70 students. How many different vote counts are possible if some student(s) do not vote.

 

I still do not understand this question.  

You explanation does not make sense to me. Just for starters the question clearly states that 1 or more students DO NOT vote. So it is impossible for 70 students to vote.

If there are 70 students and some do not vote then there can be a maximum of 69 possible votes to count.

Each of the three students can get between 0 and 69 votes, the sum of the 3 votes is obviously 69 or less.

 

I do not suppose that this is what the question means but it is impossible to know what the intended meaning is.

 Mar 7, 2023
edited by Melody  Mar 7, 2023
 #5
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2. In how many ways can we distribute 7 pieces of identical taffy and 8 pieces of identical licorice to 5 kids such that each kid recieves exactly 3 pieces of candy.

 

There are 15 peices of candy of 2 types 

I am only going to concern myself with the 7 pieces of Taffy because for each taffy allocation the licorice allocation is automatically set.

 

 Mar 7, 2023

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