MHB Sara's questions at Yahoo Answers regarding permutations and combinations

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The discussion centers on solving two combinatorial problems related to permutations and combinations. The first question involves assigning 30 teachers to 6 schools, requiring each school to receive 5 teachers, leading to the calculation of approximately 8.88 x 10^19 ways. The second question asks for the selection of 12 jurors and 3 alternates from 25 prospective jurors, which results in 1,487,285,800 combinations. Detailed explanations of the calculations for both problems are provided, illustrating the application of factorials and binomial coefficients. The thread emphasizes the importance of understanding the underlying mathematical principles for accurate problem-solving.
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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