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stuck

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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?

Oct 13, 2019

#1
+685
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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

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 ^-^

Oct 13, 2019
edited by Nirvana  Oct 13, 2019
edited by Nirvana  Oct 13, 2019
#2
0

It still didn't work

Oct 13, 2019
#3
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deleted

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Oct 13, 2019
edited by Rom  Oct 14, 2019
#4
+1914
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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

Oct 14, 2019
#5
+110206
+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?

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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.

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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.

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Oct 14, 2019
edited by Melody  Oct 14, 2019