Sara's questions at Yahoo Answers regarding permutations and combinations

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SUMMARY

The discussion centers on solving permutation and combination problems, specifically how to assign 30 teachers to 6 schools and select jurors from a group. The first problem yields a solution of approximately 8.88 x 10^19 ways to assign teachers, calculated using the formula N = 30! / (5!)^6. The second problem involves selecting 15 jurors from 25 and 3 alternates from the remaining 15, resulting in 1,487,285,800 combinations using the binomial coefficient formula.

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MarkFL
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Here are the questions:

How to solve this question about permutations? How do you get to this answer? 10 pts for best answer! :)?

Please show how you arrived at the answer (I want a detailed process explanation!)
Here's the answer to the first question: 8.88 x 10^ (19)
But how do you arrive at this answer?

Question:
In how many ways can 30 teachers be assigned to 6 schools, with each school receiving an equal number of teachers?And if you're up to the challenge ... :
Q: In how many ways can 12 jurors and 3 alternate jurors be selected from a group of 25 prospective jurors?

Thanks a bunch!

I have posted a link there to this topic so the OP can see my work.
 
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Hello Sara,

1.) Obviously we are going to have to send 5 teachers to each of the 6 schools. There are $30!$ ways to order the 30 teachers, but we have to account for the fact that for each of the 6 groups of 5 teachers, there are 5! ways to order them, and since order does not matter, we find the number of ways $N$ to do this is:

$$N=\frac{30!}{(5!)^6}=88832646059788350720\approx8.88\times10^{19}$$

2.) We need to compute the number of ways to choose 15 from 25, and multiply this with the number of ways to choose 3 from 15, hence:

$$N={25 \choose 15}{15 \choose 3}=1487285800$$
 

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