Math Combination Help: Finding Possible Divisions for New Teachers Among Schools

In summary, the problem at hand involves dividing 8 new teachers among 4 schools, with a maximum of 3 schools per teacher and a staff limit of 4 new teachers at 3 of the schools. Without the staff limit, the number of possible divisions is 4^8 + 12^8 + 24^8. However, this does not account for the possibility of some teachers teaching at one school and others teaching at two or three schools. To account for this, we must consider that each teacher has 4 choices for the school at which they do not teach, resulting in a total of 48 possibilities. With the staff limit, the number of divisions is given by the formula \sum_{0 \leq
  • #1
yitriana
36
0

Homework Statement


a) 8 new teachers are to be divided among 4 schools, and each teacher can teach at maximum 3 schools. There is a staff limit such that 3 of the schools only allow 4 new teachers. How many divisions are possible?

The Attempt at a Solution


Without staff limit
a) So if all teachers can teach at max one school, # divisions is 4^8.

If a teacher must teach at exactly two schools, this includes the possibility of all 8 teachers teaching at one school and another and no teachers at the remaining 2 schools, as well as the possibility of an evenly distributed 4 teachers for each of the four schools. A teacher has 4*3 possibilities for their two schools, so combinations is 12^8

Event of all teachers teaching at max one school and event of all teachers teaching at exactly two schools are mutually exclusive events. Teachers teaching at exactly 3 schools is (4*3*2)^8 = 24^8.
Thus, answer for part a) is 4^8 + 12^8 + 24^8

Adding staff limit
Adding staff limit for the case of teacher teaching at exactly one school, find number of division amongst all schools for case of teacher teaching at one school.

[tex]
\sum_{0 \leq m\leq 8}\sum_{\substack{i+j+k=m\\ i\leq4, j\leq4, k\leq4}}\frac{m!}{i!j!k!} \sum_{0\leq n\leq8-m} \binom{8-m}{n}
[/tex]

This is because a 0 to 8 teachers can be divided among 3 schools with staff limit (i, j and k), and depending on the number of teachers left after dividing, 8-m, consider all division of teachers from choosing 0 to 8 teachers for the 4th school with no staff limit.

Now, the hard part for me is to figure out how to apply this reasoning for the whole problem which is that teachers can teach at most three schools.

Does anyone have any suggestions?
 
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  • #2


Let's start with the without staff limit section. First, you must clarify the question. Does every teacher need to teach at a school, or can a teacher choose not to teach at all?

If every teacher must teach at one school, the number of possibilities is 48. If every teacher teaches at exactly two schools, then every teacher gets to choose 2 of the 4 schools. Does order matter? I don't agree that each teacher has 12 choices. Likewise for the case of all teachers teaching at 3 schools.

Here is an illuminating way to think about this. A teacher who teaches at one school has 4 choices for the school at which he or she teaches. A teacher who teaches at three schools has 4 choices for the school at which he or she does not teach.

Furthermore, it is true that all teachers teaching at one school, all teachers teaching at two schools, and all teachers teaching at three schools are pairwise mutually exclusive. However, they do not account for all the possibilities. What if some teachers decide to teach at one school, but others decide to teach at two schools?

How can you account for this possibility? One way is to work off the following observation: for a single teacher, teaching at one school, two schools, or three schools are pairwise mutually exclusive possibilities. More importantly, they are all the possibilities (if the answer to the first question in my post is that every teacher needs to choose at least one school).
 

1. What are math combinations?

Math combinations refer to the different ways that a set of objects can be selected or arranged. They are often used in probability and statistics to calculate the likelihood of certain events occurring.

2. How do I calculate combinations?

To calculate combinations, you can use the formula nCr = n!/r!(n-r)!, where n represents the total number of objects and r represents the number of objects being selected. Alternatively, you can use a combination calculator or table to find the number of combinations for a specific scenario.

3. What is the difference between combinations and permutations?

Combinations and permutations both involve selecting or arranging objects, but the key difference is that combinations do not take into account the order of the objects, while permutations do. In other words, combinations focus on the selection of objects without considering their arrangement, while permutations consider both selection and arrangement.

4. Can I use combinations in real-life situations?

Yes, combinations can be used in many real-life situations, such as in probability and statistics, in business to calculate different combinations of products or services, and in sports to determine different combinations of players on a team.

5. What are some common applications of combinations in science?

Combinations are commonly used in science to calculate the probability of certain outcomes in experiments or studies, to determine the different combinations of genes in genetics, and in chemistry to calculate the number of possible arrangements of molecules or atoms in a compound.

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