1. Couting and probability

Suppose that I have 13 different spices in matching canisters and two spice racks.

The first spice rack is a square that holds 8 canisters. It has one hole in each corner and one hole in the middle of each side for spices. The whole square freely rotates and there are no distinguishing marks anywhere to differentiate one side or corner from another.

The second rack is a circle with 5 wholes evenly spaced around the perimeter. Again it can spin freely and there are no distinguishing marks to differentiate one whole from any other.

In how many different ways can I put my 13 spices into the two racks?

I need an explanation how to get to this answer? Thank you for the much needed help

2. Hello, Tony!

Suppose that I have 13 different spices in matching canisters and two spice racks.

The first spice rack is a square that holds 8 canisters.
It has one hole in each corner and one hole in the middle of each side for spices.
The whole square freely rotates and there are no distinguishing marks anywhere
to differentiate one side or corner from another.

The second rack is a circle with 5 holes evenly spaced around the perimeter.
Again it can spin freely and there are no distinguishing marks
to differentiate one hole from any other.

In how many different ways can I put my 13 spices into the two racks?
First, we'll fill the square rack.

Choose 8 of the 13 spices.
. . There are: . ${13\choose8} \:=\:1287$ choices.

If we place them in a row, there would be: . $8!\:=\:40,320$ ways.

Placed in a square, there are 4 "rotations" which are identical arrangements. .**

. . Then there are: . $\frac{40,320}{4} \:=\:10,080$ ways.

Hence, there are: . $1287 \times 10,080 \:=\:12,972,960$ ways to fill the square rack.

To place the remaining 5 spices in the circular rack, there are: . $4! \:=\:24$ ways.

Therefore, there are: . $12,972,960 \times 24 \:=\: {\bf{\color{blue}311,351,040}}$ ways to fill the two racks.

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**

In our list of $8!$ possible arrangements,

. . we would have, for example: . $\boxed{\begin{array}{ccc}A & B & C \\ H & & D \\ G & F & E \end{array}}$

But we'd also have: . $\boxed{\begin{array}{ccc}G & H & A \\ F & & B \\ E & D & C\end{array}}\quad\boxed{\begin{array}{ccc}E & F & G \\ D & & H \\ C & B & A\end{array}}\quad\boxed{\begin{array}{ccc}C&D&E\\ B&&F\\A&H&G\end{array}}$

Since the rack rotates, these arrangements are identical.

3. Thank you

You did a fantastic job at explaining this answer. Much better job than even my TA could imagine doing. He thinks since he knows it so well we should all know it too. Thank you so much