Difference between revisions of "2019 AMC 10A Problems/Problem 17"
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==Solution 3 (Educated Guess & MOST EFFICIENT)== | ==Solution 3 (Educated Guess & MOST EFFICIENT)== | ||
| − | We start by doing the casework for Case 1, as stated in Solution 2. We see that it would be = <math>280</math>. This eliminates answer choices A, B, and C since they are all to small. Then we see the answer choice E is way to big. This makes us choose <math>1,260</math>. | + | We start by doing the casework for Case 1, as stated in Solution 2. We see that it would be = <math>280</math>. This eliminates answer choices A, B, and C since they are all to small. Then we see the answer choice E is way to big. This makes us choose <math>1,260</math>. boxed{\textbf{(D) } 1,260} |
~ Continuous_Pi | ~ Continuous_Pi | ||
Revision as of 21:30, 25 May 2024
Contents
Problem
A child builds towers using identically shaped cubes of different colors. How many different towers with a height 
cubes can the child build with 
red cubes, 
blue cubes, and 
green cubes? (One cube will be left out.)
Solution 1 (Clever Bijection)
Arranging eight cubes is the same as arranging the nine cubes first, and then removing the last cube. In other words, there is a one-to-one correspondence between every arrangement of nine cubes, and every actual valid arrangement. Thus, we initially get 
. However, we have overcounted, because the red cubes can be permuted to have the same overall arrangement, and the same applies with the blue and green cubes. Thus, we have to divide by the 
ways to arrange the red cubes, the 
ways to arrange the blue cubes, and the 
ways to arrange the green cubes. Thus we have 
different possible towers.
Note:
this can be written more compactly as
![\[\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}\]](http://latex.artofproblemsolving.com/6/d/a/6da10c0235cb17584bf12c617c5d555d5a59bf41.png)
Solution 2 (Casework)
We can divide the problem into three cases, each representing one cube to be excluded:
Case 1: The red cube is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The number of possible arrangements is 
. Note that we do not need to multiply by the number of red cubes because there is no way to distinguish between the first red cube and the second.
Case 2: The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The number of possible arrangements is 
.
Case 3: The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The number of possible arrangements is 
.
Adding up the individual cases from above gives the answer as 
.
Solution 3 (Educated Guess & MOST EFFICIENT)
We start by doing the casework for Case 1, as stated in Solution 2. We see that it would be = 
. This eliminates answer choices A, B, and C since they are all to small. Then we see the answer choice E is way to big. This makes us choose 
. boxed{\textbf{(D) } 1,260}
~ Continuous_Pi
Video Solution by OmegaLearn
https://youtu.be/0W3VmFp55cM?t=612
~ pi_is_3.14
Video Solution
https://youtu.be/3MiGotKnC_U?t=1083 ~ ThePuzzlr
Video Solution
Education, the Study of Everything
Video Solution
https://youtu.be/D9RT489UwyQ?t=1338
~ AMBRIGGS
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
