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The question is:

A baby has nine different toy animals. Five of them are red and four of them are blue. She arranges them in a line so that the colours are arranged symmetrically. How many different arrangements are possible?

I understand that they key here is that they must be arranged symmetrically. Given the unequal numbers of red to blue, isn't the only arrangement possible one being B B R R R R R B B?

So shouldn't it be $4! * 5!$ ?

However, the answer is $6 *5 !* 4! $

Which combinations did I miss, or is it a more straightforward way of doing this?

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up vote 1 down vote accepted

Possible steps are deciding arrangements of
1. colors(red/blue, assuming toys of same color is same)
2. red toy animals
3. blue toy animals

  1. Since it should be arranged symmetrically, red should be at center(5th) and each side should have 2 reds and 2 blues. And right side will be decided if left side is decided. Therefore, ${ 4 \choose 2 }= 6$.
  2. There is five red toy animals(including center). Therefore, $5!$.
  3. There is four blue toy animals. Therefore, $4!$.

Therefore, $6 \cdot 5! \cdot 4!$.

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$5! = $ number of possible arrangements of red animals

$4! = $ number of possible arrangements of blue animals

Now for symmetry you need arrangement $x_1x_2x_3x_4Rx_4x_3x_2x_1$. The number of possible arrangements like this is equal to ${ 4 \choose 2 }= 6$ (among $x_1x_2x_3x_4$ we choose two x's to be blue). The answer is $6 \cdot 5! \cdot 4!$.

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Hint also R B R B R B R B R is also possible and key is to keep red coloured toy at the $5th$ place to keep symmetry in tact

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