2019 AMC 10A Problem 17

Step 1: a trick to handle the left-out cube.

1. Take any tower of $8$ cubes. Exactly one cube is left over, and you know its color. Put it on top: you get a stack of all $9$ cubes.
2. Going the other way, take any stack of all $9$ cubes and remove the top cube: you get a tower of $8$.
3. These two moves undo each other, so each $8$-tower matches exactly one $9$-stack. The two counts are equal.

Step 2: count the $9$-cube stacks.

4. If all $9$ cubes looked different, there would be $9!$ orders.
5. But the $2$ reds look alike. Swapping them gives the same stack, so divide by $2!$. In the same way, divide by $3!$ for the blues and $4!$ for the greens.
$$\frac{9!}{2!\,3!\,4!} = \frac{362880}{2 \cdot 6 \cdot 24} = \frac{362880}{288} = 1260.$$

The answer is $\boxed{\textbf{(D)}\ 1{,}260}$.

When some objects are identical, count all orders as if they were different, then divide by the number of ways to shuffle each group of look-alikes. The "one cube left out" twist disappears once you notice the leftover cube can always be placed on top, which turns towers of $8$ into stacks of $9$ one-for-one.

Split into cases by which color is left out.

• Leave out a red: $1$ red, $3$ blue, $4$ green, so $\frac{8!}{1!\,3!\,4!} = 280$ towers.
• Leave out a blue: $2$ red, $2$ blue, $4$ green, so $\frac{8!}{2!\,2!\,4!} = 420$ towers.
• Leave out a green: $2$ red, $3$ blue, $3$ green, so $\frac{8!}{2!\,3!\,3!} = 560$ towers.

Total: $280 + 420 + 560 = 1260$.