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Permutation: How many ways can you seat 8 students


phxmba

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I saw this in one of the websites

 

1. in how many ways can you seat 8 students in a row of 8 chairs ?

answer: 8!

 

2. in how many ways can you seat 8 students in a row of 9 chairs ?

answer: 9!

 

3. in how many ways can you seat 8 students in a row of 10 chairs ?

answer: 10! / 2!

 

4. in how many ways can you seat 8 students in a ring of 8 chairs ?

answer: 7!

 

I wasn't sure about the answer to questions 2 & 3 - how they are different. Can someone explain?

 

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You should take into account that seats without poeple are the same.

If you don't consider this,the answer is 10!,but when you change the two free seats,you can figure out that the array isn't changed,so 10!/2!

 

Furthermore,in the problem below,

in how many ways can you seat 8 students in a row of 11 chairs ?

the answer is 11!/3!

clear?

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I will interpret it in such a way:

Assuming that there is an apple on one free seat , we can change this problem to "in how many ways can you seat 10 students in a row of 10 chairs ? ",thus the answer is 10!.Then,I take away the apple,the situation changes into "in how many ways can you seat 8 students in a row of 10 chairs ? ".We assume there are X ways.

I put the former apple on one of the free seats.I have 2! chooses.So I can obtain an equation:2!*X=10!,X=10!/2!.

Clear?Come on.

 

 

 

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Thanks Phase. Also, I got another explanation to the problem.

 

Ways of seating 8 students in 8 seats - 8P8 -> 8!

 

Ways of seating 8 students in 9 seats (selecting 8 out of available 9) - 9P8 -> 9!/1!

 

Ways of seating 8 students in 10 seats (selecting 8 out of available 10) - 10P8 -> 10!/2!

 

 

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