# Boys and Girls

• Nov 1st 2009, 12:10 PM
Aquafina
Boys and Girls
3 girls and 4 boys were standing in a circle. What is the probability that two girls are standing together, but the third one is not next to them?
• Nov 1st 2009, 06:37 PM
ANDS!
Question doesn't make any sense. One what is not with them?
• Nov 1st 2009, 09:30 PM
Aquafina
Quote:

Originally Posted by ANDS!
Question doesn't make any sense. One what is not with them?

I have edited it to be clearer. This is what I did, there are 7 arrangements that 2 girls can stand in:

GG***** the other girl can stand in 4 different places here
*GG**** other girl in 3 places
**GG*** 3
***GG** 3
****GG* 3
*****GG 4

So in total: 2x(4+3+3) = 20

Is this correct? Do you have any other methods that you could show?
• Nov 1st 2009, 09:35 PM
Robb
Quote:

Originally Posted by Aquafina
I have edited it to be clearer. This is what I did, there are 7 arrangements that 2 girls can stand in:

GG***** the other girl can stand in 4 different places here
*GG**** other girl in 3 places
**GG*** 3
***GG** 3
****GG* 3
*****GG 4

So in total: 2x(4+3+3) = 20

Is this correct? Do you have any other methods that you could show?

Dont forget that they are in a circle, so in your first case the third girl cannot take the position of the last astrix, as she would then be next to the first girl
• Nov 1st 2009, 09:39 PM
Aquafina
Quote:

Originally Posted by Robb
Dont forget that they are in a circle, so in your first case the third girl cannot take the position of the last astrix, as she would then be next to the first girl

Ah thank you, so it should be:

GG***** 3
*GG**** 3
**GG*** 3
***GG** 3
****GG* 3
*****GG 3

so 18 in total?

Do you have any other methods?
• Nov 4th 2009, 11:30 AM
jt5271
probability = no. ways you can arrange two girls together without third being adjacent / total no ways to arrange 3 girls and 4 boys in circle.

B- boy
G-girl
BGGB - C

First part is synomymous to arranging B,B,G,C on a straight line = 4!/2! = 12

Second part = 7!/3!4! = 35

Prob = 12/35