Sara's questions at Yahoo Answers regarding permutations and combinations

In summary, the first question is asking for the number of ways to assign 30 teachers to 6 schools with each school receiving an equal number of teachers, which is approximately 8.88 x 10^19. The second question is asking for the number of ways to select 12 jurors and 3 alternate jurors from a group of 25 prospective jurors, which is 1487285800.
  • #1
MarkFL
Gold Member
MHB
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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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  • #2
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:

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

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:

\(\displaystyle N={25 \choose 15}{15 \choose 3}=1487285800\)
 

Related to Sara's questions at Yahoo Answers regarding permutations and combinations

1. What is the difference between permutations and combinations?

Permutations and combinations are both ways of arranging or selecting items from a group. The main difference is that permutations take into account the order of the items, while combinations do not. In other words, with permutations, the order of the items matters, but with combinations, it does not.

2. How do I know when to use permutations and when to use combinations?

The key factor in determining whether to use permutations or combinations is whether the order of the items matters in your situation. If the order does not matter, use combinations. If the order does matter, use permutations.

3. How do I calculate permutations and combinations?

The formula for calculating permutations is n!/(n-r)!, where n is the total number of items and r is the number of items being selected. The formula for combinations is n!/r!(n-r)!. Alternatively, you can use a calculator or computer program to quickly calculate permutations and combinations.

4. Can permutations and combinations be used in real-life situations?

Yes, permutations and combinations are used in many real-life situations, such as in probability and statistics, gambling, and in everyday situations like choosing a lottery ticket or arranging a seating chart.

5. What are some common misconceptions about permutations and combinations?

One common misconception is that permutations and combinations are the same thing. As mentioned before, they are different because of the consideration of order. Another misconception is that permutations and combinations are only used in math and have no practical applications. In reality, they are used in various fields and can be helpful in solving many problems.

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