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I am really stuck on this question....

Calculate the number of ways in which three girls and four boys can be seated on a row of seven chairs if each arrangement is to be symmetrical?

The answer is 432, no idea how they got it?
Reply 1
Avatar for vc94
vc94
12
BBGGGBB or BGBGBGB are symmetric...can you count the no. of possibilities?
Reply 2
Avatar for Galadirith
Galadirith
Hi kawilliam,

Ok well lets have a think about this. Now we require a symmetrical row of 3 girls and 4 boys. Now think about this, what would happen if a girl was NOT at the center of the row ie NOT on the 4th seat. Would it be possible to have a symmetrical arrangment if this is so?

Answering that, now, if we consider for now all boys to the same and all girls to be the same, we would describe this then as a 'Combination' which I'm sure you know ;D, what are the possible symmetric arrangments. Hint, try thinking only about the arragment of the first 3 or last 3: (don't look untill you have found them for yourself, note that their are more than 2 ;D)

Spoiler


Ok we now have the possible 'Combinations' of the boys and girls in a symmetrical combination. However, we are not actually looking for the possible symmetrical combinations, we are looking for the number of possible 'permutations' or arrangements, meaning that each boy and girl are actually considered to be different or distinct from one another.

So this means that each boy can potential sit in any of the 4 seats for boys in any of the combination, and each girl can potential sit in any of the 3 seats for girls in any of the combinations. Take time to take that it, made my brain hurt just writing it :biggrin:. Now have a real think about how you can use your knowledge of permutations to find the answer :biggrin:
If we now consider any one of the combinations of the boys and girls :

BBGGGBB
BGBGBGB
GBBGBBG

Let's take the first one for example:
B B G G G B B

We have 4 boys and 3 girls, so we'd have
= 4 x 3 x ( 3 x 2 x 1 ) x 2 x 1
= 4 x 3 x 3! x 2 x 1
= 4! x 3! ways (or if you want 4P4 x 3P3)

Now the other 2 combinations give the same number of ways

Let's take a second one
B G B G B G B

We'd have
4 x 3 x 3 x 2 x 2 x 1 x 1
which is the same as 4! x 3!


As we have 3 combinations
BBGGGBB or BGBGBGB or GBBGBBG

We'd have 4! x 3! + 4! x 3! + 4! x 3!
= 3 x (4! × 3!)
= 432 ways
(edited 3 years ago)

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