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.

i posted this question earlier but I was unable to receive an answer.

Guest Mar 5, 2023

#1**0 **

1.

Let's call the number of votes for the first student A, the number of votes for the second student B, and the number of votes for the third student C. The total number of votes cast must be less than or equal to 70.

If all 70 students vote, then the number of possible vote counts is the number of ways to distribute 70 votes among the three candidates, which is given by the stars and bars formula:

N = (70 + 3 - 1) choose (3 - 1) = 72 choose 2 = 2,556

However, we also need to consider the cases where some students do not vote. Suppose k students do not vote, where 0 <= k <= 70. Then the number of possible vote counts is the number of ways to distribute the remaining 70 - k votes among the three candidates, which is again given by the stars and bars formula:

Nk = (70 - k + 3 - 1) choose (3 - 1) = (72 - k) choose 2

So the total number of possible vote counts is:

Ntotal = N0 + N1 + N2 + ... + N70 = (72 choose 2) + (71 choose 2) + (70 choose 2) + ... + (2 choose 2) = (1/3) * (72 choose 3) = 8,296

Therefore, there are 8,296 different possible vote counts if some students do not vote.

Guest Mar 5, 2023

#2**0 **

For the second problem

First, let's consider distributing the taffy. We need to distribute 7 identical pieces of taffy to 5 kids such that each kid receives exactly 3 pieces. This is equivalent to partitioning 7 into 5 non-negative parts, where each part represents the number of taffy pieces received by a kid. By the stars and bars formula, the number of ways to partition 7 into 5 non-negative parts is:

${7+5-1 \choose 5-1}={11 \choose 4}=330$

Now, we need to distribute the 8 identical pieces of licorice among the 5 kids such that each kid receives exactly 3 pieces. Since each kid has already received 3 pieces of taffy, we only need to distribute 8 - 5(3) = -7 pieces of licorice, which means some kids will not receive any licorice.

There are two cases to consider:

Case 1: Two kids receive all the licorice, and the other three kids receive no licorice. There are ${5 \choose 2}$ ways to choose which two kids receive all the licorice. Then, each of the two chosen kids will receive 3 pieces of taffy and 4 pieces of licorice, while the other three kids will each receive 3 pieces of taffy and no licorice. Thus, there are ${5 \choose 2}$ ways to distribute the candy in this case.

Case 2: One kid receives all the licorice, and two other kids receive one piece of licorice each. There are ${5 \choose 1}$ ways to choose which kid receives all the licorice. Then, the chosen kid will receive 3 pieces of taffy and 5 pieces of licorice, while the other two kids who receive 1 piece of licorice each will receive 3 pieces of taffy and 1 piece of licorice each. There are ${4 \choose 2}$ ways to choose which two kids receive the single pieces of licorice. Thus, there are ${5 \choose 1} \times {4 \choose 2}$ ways to distribute the candy in this case.

Therefore, the total number of ways to distribute the candy is ${5 \choose 2} + {5 \choose 1} \times {4 \choose 2} = 10 + 30 = 40$.

Therefore, there are 330 × 40 = 13200 ways to distribute the taffy and licorice to the 5 kids.

Guest Mar 5, 2023