+0  
 
0
1563
1
avatar

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.)

Guest Mar 14, 2015

Best Answer 

 #1
avatar+91513 
+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,

ABCD = ADCB

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

ABCD = BCDA = CDAB=DABC   now counter clockwise   = DCBA=ADCB=BADC=CBAD   8 all the same

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

ADBC       8 the same

ADCB       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
Sort: 

1+0 Answers

 #1
avatar+91513 
+5
Best Answer

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,

ABCD = ADCB

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

ABCD = BCDA = CDAB=DABC   now counter clockwise   = DCBA=ADCB=BADC=CBAD   8 all the same

ABDC       8 the same

ACBD       8 the same

ACDB       8 the same

ADBC       8 the same

ADCB       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

16 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details