Question

A child builds towers using identically shaped cubes of different colors. How many different towers with a height $8$ cubes can the child build with $2$ red cubes, $3$ blue cubes, and $4$ green cubes? (One cube will be left out.)

$24$

$288$

$312$

$1,260$

$40,320$

Solution:

Every tower of height $8$ could have been formed by creating a tower of height $9$ and removing the top cube.

This shows that there is a one-to-one correspondence between towers of height $8$ and $9.$

There are $9!$ ways to make a tower of height $9,$ but we are overcounting since there are multiple cubes of the same color.

We have to divide through by $2!$ ways to arrange the red cubes, $3!$ for the blue cubes, and $4!$ for the green cubes.

Therefore, the number of valid arrangements is $\dfrac{9!}{2! \cdot 3! \cdot 4!} = 1,260.$

Thus, D is the correct answer.

2019 AMC 10A Problem 17

Solution:

Problem 17 in Other Years