Circular permutations with two tables? (2 Viewers)

[fan96]

e.g. How many ways can 8 people be seated at two tables with 5 and 3 seats each?

How do I approach this problem?

Also, my teacher used this question as an example to say that we should always learn the formula as n!/n and not (n-1)!. I don't understand that, can anyone figure out why he said that?

 

[pikachu975]

You would have to pick 5 people for the first table so 8C5, and arrange them via 4!. Then arrange the remaining 3 using 2!. Honestly idk what your teacher is saying as n!/n = n(n-1)!/n = (n-1)!....

So the answer is 8C5 x 4! x 2!

And it works by picking 3 for the first table as 8C3 = 8C5

 

[leehuan]

You would have to pick 5 people for the first table so 8C5, and arrange them via 4!. Then arrange the remaining 3 using 2!. Honestly idk what your teacher is saying as n!/n = n(n-1)!/n = (n-1)!....

It's a bit lame but it's not unjustified, because a lot of people rote-learn the (n-1)! formula without understanding why it works. Thus they can't adapt it to other circular arrangements (e.g. those, but two must sit next to each other).

 

[InteGrand]

e.g. How many ways can 8 people be seated at two tables with 5 and 3 seats each?

How do I approach this problem?

Also, my teacher used this question as an example to say that we should always learn the formula as n!/n and not (n-1)!. I don't understand that, can anyone figure out why he said that?

You should be able to use either formula, provided you understand why it's true.

(You can even use it if you don't understand why it's true, but this is not recommended.)

 

[fan96]

You would have to pick 5 people for the first table so 8C5, and arrange them via 4!. Then arrange the remaining 3 using 2!. Honestly idk what your teacher is saying as n!/n = n(n-1)!/n = (n-1)!....

So the answer is 8C5 x 4! x 2!

And it works by picking 3 for the first table as 8C3 = 8C5

Thanks for the answer.

I notice now that you can get the same answer by doing 8P5 / 5 for one table (since 8P5 gives you the number of possible arrangements in a straight line, which can be divided by 5 seats to get the number of arrangements for a circle) multiplied by 2! (or 3!/3) for the other table.

I think I understand my teacher was referring to now, since someone who learns the formula as (n-1)! wouldn't be able to apply it properly to the question.