Combinatorics/Permutation problem?

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SUMMARY

The problem involves seating arrangements for 4 American, 2 French, and 3 British individuals, with the constraint that individuals of the same nationality must sit together. By treating each nationality as a single unit, the total arrangements can be calculated as 3! for the groups and 4!, 2!, and 3! for the arrangements within each group. The final calculation yields a total of 1728 unique seating arrangements, confirmed by multiple participants in the discussion.

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Homework Statement



If 4 American, 2 French, and 3 British people are to be seated in a row, how many seating arrangements are possible when people of the same nationality must sit next to each other?

Homework Equations





The Attempt at a Solution



If I "glue" the people together like so: ABF then there are 3! ways to arrange them. There are 4! ways to arrange the Americans, 2! to arrange the French, and 3! ways to arrange the british. Is the answer 3!4!2!3!= 1728?
 
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I thought you had run away! :smile:

I'm starting to think that you like these problems!


(Oh, and yes, you have it right! :wink:)
 


It's a miracle!

Oh, and I don't like these problems, I like passing my classes.
 

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