+0

In how many ways can 4 boys and 4 girls sit around a circle table if all the boys sit together? (Rotations of the same arrangement are still

0
2389
1

In how many ways can 4 boys and 4 girls sit around a circle table if all the boys sit together? (Rotations of the same arrangement are still considered the same. Each boy and girl is unique, not interchangeable.)

Mar 14, 2015

#1
+5

3!*4!   but I think this should be divided by 2 because each clockwise possibilty has an anticlockwise possiblility tht would be classed as the same.

so I think

$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$

---------------------------------------------------------

this is something we have been debating today so I am going to try and look at an easier question and see what happens.

I am going to have just 4 children sitting in a circle   they will be ABC and D

no restrictions except that clockwise is considered the same as anticlockwise so, fo instance,

I think there should be 3!/2  ways = 3 ways  Mmm

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

That is 6 choices starting with A, there will be another 6 starting with B, 6 with C and 6 with D

that is 6*4=24 permutations.

BUT how many of these are really the same?

24 divided by 8 = 3

I have colour coded to show the 3 distinct possibilities

-------------------------------------------

SO I AM STICKING WITH MY ORIGINAL ANSWER      ლ(o◡oლ)

Mar 15, 2015

#1
+5

3!*4!   but I think this should be divided by 2 because each clockwise possibilty has an anticlockwise possiblility tht would be classed as the same.

so I think

$${\frac{{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!}}{{\mathtt{2}}}} = {\mathtt{72}}$$

---------------------------------------------------------

this is something we have been debating today so I am going to try and look at an easier question and see what happens.

I am going to have just 4 children sitting in a circle   they will be ABC and D

no restrictions except that clockwise is considered the same as anticlockwise so, fo instance,

I think there should be 3!/2  ways = 3 ways  Mmm

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

That is 6 choices starting with A, there will be another 6 starting with B, 6 with C and 6 with D

that is 6*4=24 permutations.

BUT how many of these are really the same?

24 divided by 8 = 3

I have colour coded to show the 3 distinct possibilities

-------------------------------------------

SO I AM STICKING WITH MY ORIGINAL ANSWER      ლ(o◡oლ)

Melody Mar 15, 2015