At a birthday party, 5 boys and 3 girls are seated around a table. How many different arrangements are possible if no two girls are seated next to each other?

Guest Oct 13, 2019

#1**+1 **

hmm i remember a couple of years ago i got something similar to this on one of my homeworks and i found a youtube video that explained it pretty well...

let me see if i can find it

https://www.youtube.com/watch?v=3dQLWb86U18

it's not the same exact problem but it is very very similar to the one you need help on...

i hope it helps you ^-^

Nirvana Oct 13, 2019

#4**+3 **

**Solution**:

\(\text {There are $\mathbf{(5-1)!}$ ways to arrange the boys in a circle. } | \small \text {(circular permutation) }\\ \text {There are five $\mathbf{(5)}$ positions between the boys. }\\ \text {There are $\mathbf{P(5,3)}$ ways to arrange the girls to occupy the positions between the boys.}\\ \mathbf{(4!) * P(5,3) = }\\ \mathbf{(24) * (60) = 1440} \\ \text{ }\\ \large \text {There are $\mathbf{1440}$ unique circular arrangements of 5 boys and 3 girls,} \\ \large \text {where each girl is separated from the other girls by at least one boy. }\\ \)

GA

GingerAle Oct 14, 2019

#5**+2 **

At a birthday party, 5 boys and 3 girls are seated around a table.

How many different arrangements are possible if no two girls are seated next to each other?

Normally for these questions rotations are considered to be the same.

Reflections are considered to be different

BUT

Are you interested in gender arangements or individual people arrangements?

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

If you are **only** interested in gender positions then.

Seat a girl first. She is like the begining of the row.

Now there is 5 boys

GBBBBB

We need to slot in 2 more girls

GB*B*B*B*B

The stars indicate where the other 2 girls can go.

4C2 = **6 gender arrangements** are possible.

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

If you are interested in the position of individuals then

Stil sit any girl first. (could have chosen a boy if you wanted to)

Now there is 5 boys and 2 girls left to seat.

Sit the 5 boys first. This can be done in 5! = 120 ways

So we have GBBBBB and we have to sit 2 more girls.

GB*B*B*B*B

4C2*2 = 12 ways.

So that is 120*12 = **1440 individual child arrangements. **

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

Melody Oct 14, 2019