In how many ways can 4 boys and 4 girls sit around a circle table if each boy sits directly between two girls? (Rotations of the same arrangement are still considered the same. Each boy and girl is unique, not interchangeable.)
+118723 You mean you posted the same question twice? :)) I whited the other one. Don't sweat it. :)
+118723 this may not be right but here goes.
I am going to sit a boy down - anywhere.
there are 3 places for the other 3 boys that is 3!
and 4 places for the girls that is 4!
but you probably have to cut that in half to cover the clockwise anticlockwise condition.
so maybe it is
$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$
I'm sorry, but when I put this answer, it was wrong. I can't explain because I don't know
I don't see a way to accomplish this since there will always be one pair of boys next to each other.
+130514 I'll try this one.....here goes nothing......!!!
We have 4 ways of choosing a boy to sit in any of the chairs.
And we have 4 ways to select any girl who sits to the right of him......3 ways to seat the next boy to the right of her......3 ways to seat the next girl to the right of the second boy, etc.
So....the total number of ways is just 4! x 4! = 576 ways
Don't know if I'm correct, but that's my two cents worth.......
+130514 Disregard my previous answer....I want to think about this one a little more....!!!
+130514 Choose a boy to "anchor" the circle......he can sit in any seat
To his right, any of the 4 girls can be chosen
To her right, any of the 3 other boys can be chosen
Etc.
So..... the total ways = 4! x 3! = 24 x 6 = 144 ways
?????
+118723 Yes Chris but what about the clockwise/anticlockwise clause?
Each clockwise permutation has a matching anticlockwise permutation so i think my first answer is correct. :/
+130514 I'm not sure as to what the correct answer might be......yours may be good.......!!!
