Explain method behind permutations question (high school level)

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.


I cannot explain your thinking, because $4!$ simply doesn't count the question. Further, adding one in a scenario like this isn't something that is generally ever done.

Although not explicitly stated, the problem seems to imply that the toys are themselves distinguishable and that the symmetry we are hoping for is reflective symmetry with respect to colors about the middle.

The book's (and correct) answer can be explained via multiplication principle:

  • Begin by noticing as you had that the middle spot must be red. Continue by noticing the first four spots must have exactly two of them being red and exactly two of them being blue. Once we have picked the arrangement of red and blue in the first four spots, the final four spots must mirror that selection. There are $\binom{4}{2}=6$ ways to reserve which of the slots are for red toys and which are for blue toys respecting the symmetry requirement.
  • From left to right, among the spaces reserved for red toys, pick which red toy is used. $5$ choices for the first appearing slot reserved for red. $4$ choices for the second appearing slot reserved for red... on up to only $1$ choice for the final appearing slot reserved for red, for a total of $5\cdot 4\cdot 3\cdot 2\cdot 1 = 5!$ ways to place the red toys into their reserved slots.
  • From left to right, among the spaces reserved for blue toys, pick which blue toy is used. $4$ choices for the first appearing slot reserved for blue, $3$ choices for the second appearing slot reserved for blue, etc... for a total of $4\cdot 3\cdot 2\cdot 1 = 4!$ ways to place the blue toys into their reserved slots.

Applying multiplication principle, there are $\binom{4}{2}\cdot 5!\cdot 4!$ ways to arrange the toys respecting the conditions.

Tags:

Permutations

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