There are 2015 people sitting at a table, each person belongs to one of the four group: student,professors, presidents and deans. Students hate professors, and will never sit next to them. Likewise for presidents and deans. Explain why, in every possible seating, there must be two people from each group sitting next to each other who are from the same group.
If he asked how many ways you can sit them then I would be able to answer, but to explain why I'm not sure how.
Are you sure of your problem?
2,015 people at one table??????
Yes that is what he asked.
There are 2015 people from universities across the country, all sitting around a massive table. Each person belongs to one of four groups: students, professors, deans and presidents. Students hate professors, and will never sit next to them. Likewise for presidents and deans. Explain why, in every possible seating, there must be two people from each group sitting next to each other who are from the same group.
To explain why there must be at least two people from each group sitting next to each other in every possible seating arrangement, we can use a proof by contradiction.
First, let's assume that it's possible to arrange the 2015 people such that no two people from the same group are sitting next to each other.
Since there are four groups, we know that each group must have a different number of people, let's say x, y, z, and w, where x, y, z, and w are positive integers. The sum of these four numbers must equal 2015, as there are 2015 people in total. So, we have:
x + y + z + w = 2015 ------ (Equation 1)
Now, based on the assumption that no two people from the same group can sit next to each other, we can conclude that the minimum possible distance between any two people from the same group must be at least one seat.
Let's consider the maximum number of empty seats between two people from the same group. Since there are 2015 people, the maximum number of empty seats between two people is 2013. This means that at least one seat must separate any two people from the same group.
Now, let's evaluate the number of available seats for each group based on the assumption:
- For the first group, there are x seats available, as there must be at least one seat between any two people from the first group.
- For the second group, there are y-1 seats available, as there must be at least one seat between any two people from the second group, and we've already placed one person from the first group.
- For the third group, there are z-2 seats available, following the same logic as above.
- For the fourth group, there are w-3 seats available, following the same logic as above.
Based on this, the total number of available seats can be calculated as follows:
x + (y-1) + (z-2) + (w-3) = x + y + z + w - 6 = 2015 - 6 = 2009 -------- (Equation 2)
Now, we can compare Equation 2 with Equation 1:
x + (y-1) + (z-2) + (w-3) = x + y + z + w - 6 = 2015 - 6 = 2009
x + y + z + w = 2015
We can see that Equation 2 is smaller than Equation 1. This contradicts our assumption that it's possible to arrange the 2015 people such that no two people from the same group are sitting next to each other. Therefore, our assumption is incorrect.
Hence, there must be at least two people from each group sitting next to each other in every possible seating arrangement.