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?
Best Answer
$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 $6 \cdot 5! \cdot 4!$.