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.)

$24$

$288$

$312$

$1,260$

$40,320$

Answer: D

Concepts:multiset permutations bijection

Difficulty rating: 1480

Solution:

Given a valid height- $8$ tower, its color counts determine the one unused cube; place that cube on top. Conversely, removing the top cube from any arrangement of all $9$ cubes gives a valid height- $8$ tower. These operations are inverses, so the desired towers are in one-to-one correspondence with arrangements of all $9$ cubes.

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

Answer: D

Difficulty rating: 1480

Solution:

Problem 17 in Other Years