I am in high school and got a question in my textbook which reads:
A baby has nine different toy animals. 5 of them are red and 4 of them are blue. She arranges the toys in a line so that the colours are arranged symmetrically. How many arrangements are possible?
This is how I reasoned: to fulfil the symmetrical requirement, the middle toy must be red. This leaves 8 toys, 4 on each side of the middle red toy – 4 of these are red, and 4 are blue. To calculate the permutations of the remaining 8, keeping the symmetrical requirement in mind, we calculate 4!.
Therefore, the total number of arrangements is $4! + 1 = 25$
This was not the correct answer – the correct answer is $6*5!*4! = 17280$
Could someone please explain where my thinking fails me? Thank you!
Best Answer
So, you are correct in saying that the middle toy must be red. Note that we have spaces as follows: $$ [1][2][3][4][R][6][7][8][9].$$ Because the arrangement is symmetric in color, we only care about arranging the color on one side of the $5^{th}$ slot, because this determines the coloring of the entire arrangement. For instance, if we arrange $6-9$ as $$ [R][B][R][B]$$ the overall configuration is $$ [B][R][B][R][R][R][B][R][B].$$ So, there are $$ \frac{4!}{(2!)^2}=6$$ ways to arrange the colors. Then there are $5!$ ways to permute the distinct red toys, and $4!$ ways to permute the distinct blue toys. Using the product rule we have $$ 6(5!)(4!)$$ permutations, as desired.