Five Americans, three Italians and six French sit randomly in a round table such that each sits together with their countrymen. How many seating arrangements are possible?
would it be 5!3!6!
i think that would be how it would be if they were in a row....
so maybe the two ends of the row would have the same nationality?
am i making u nervous by watching you lol
not really..no...
you do make me creeped out though....
its in a cirlce so you have to restrict..different than in a row
...but 5!3!6! is for the row right?
ahh i got it 4!3!6! + 5!2!6! + 5!3!5!
for the row?
circle
hmm no that's not it
hmm..i dont think that it is that easy.
okay so you have to fix one person first..then arrange the others
that's why i used what i wrote a while ago...
but it didnt look right..hang on..lemme try
yes it wasn't
it's too small
it's not 5!3!6! + 4!3!6! + 5!2!6! + 5!3!5! either
the number of ways to arrange n objects in a circle is (n-1)!
i think you have to do it in cases. that is correct...but your question you have to arrange the 3 nationalities hmm..
i think this is more difficult than how i make it out to be
do they alternate?
i don't think so
like AAIIFF.. or AAAIIIFFF
i now have no idea how to do this
taking each nationality group as 1 entity , those 3 entities can be arranges in 2! ways(in circle) , so wouldn't it be simply 2!(5!3!6!)
....that's right.....
what did you do?
explain what happened please @hartnn
lol! i think u understand how n people can be arranged in (n-1)! ways in circular fashion.....or u need to understand that ?
i assume it's because one can sit on the top and then (n-1)! is the permutation the others can sit on the remaining seats
on the top ?? take 3 people, how many ways can they sit in circular fashion ? 2 ? or more ?|dw:1348727561739:dw|
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