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2019 AMC 10A Problem 17

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2019 AMC 10A Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 3 of 5 · about 4 min · position P16-20

A child builds towers using identically shaped cubes of different colors. How many different towers with a height 88 cubes can the child build with 22 red cubes, 33 blue cubes, and 44 green cubes? (One cube will be left out.)

  1. A)

    2424

  2. B)

    288288

  3. C)

    312312

  4. D)

    1,2601,260

  5. E)

    40,32040,320

Answer

D

Key insight

Arrange all 9 cubes in a line, 9!/(2! 3! 4!) = 1260 ways, and remove the top one; each tower of 8 arises from exactly one such line.

Solution

Imagine stacking all 99 cubes, then removing the top one. The tower of 88 that remains is exactly the kind of tower being counted, and conversely every 88-tower comes from exactly one 99-stack (put the leftover cube on top). So the number of 88-towers equals the number of 99-stacks.

Stacks of 99 cubes with 22 red, 33 blue and 44 green identical cubes number
9!2! 3! 4!=3628802⋅6⋅24=362880288=1260.\frac{9!}{2!\,3!\,4!} = \frac{362880}{2 \cdot 6 \cdot 24} = \frac{362880}{288} = 1260.

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

Why this works

Arranging objects with repeated identical types is n!a! b! c!\frac{n!}{a!\,b!\,c!}. The "one left out" twist can be handled by casework on the omitted color, but the cleaner idea is a bijection: the leftover cube is determined by the tower, so tacking it on top pairs 88-towers with 99-stacks one-to-one.

Alternative approach

Casework on which cube is omitted: leave out a red (11 red, 33 blue, 44 green): 8!1!3!4!=280\frac{8!}{1!3!4!} = 280; a blue: 8!2!2!4!=420\frac{8!}{2!2!4!} = 420; a green: 8!2!3!3!=560\frac{8!}{2!3!3!} = 560. Total 280+420+560=1260280 + 420 + 560 = 1260.

The trap

Using 8! or 9! for identical cubes without dividing by the repeated colors, or forgetting to account for which color is left out.

Common mistakes

  • Using 8! or 9! for identical cubes without dividing by the repeated colors, or forgetting to account for which color is left out.
  • Computing only one case, such as 8!2!3!3!=560\frac{8!}{2!3!3!} = 560, or adding a factor of 33 for "which color is left out" on top of the casework.

Techniques

Map the objects to something easier to count · Split into exhaustive cases and handle each