(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

3. 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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# Homework Help: Math combinations help

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