Difference between revisions of "2019 AMC 10A Problems/Problem 17"
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(Introduced a new manner of solving the problem. I apologize for any inconveniences and mistakes in the text as I am not familiar with the typesetting, especially the equation typesetting.) |
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This is written more compactly as | This is written more compactly as | ||
<cmath>\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}</cmath> | <cmath>\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}</cmath> | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can do this on a case-by-case basis. We divide the problem into three cases, each representing one cube to be excluded | ||
| + | 1) The red cube is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The amount of arrangements would then be <math>(8!/(4!3!))=280</math>. Note that we do not need to multiply by the number of red cubes because there is no way to discriminate between the first red cube and the second. | ||
| + | 2) The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The amount of arrangements would then be <math>(8!/(2!2!4!))</math>=<math>420</math>. | ||
| + | 3) The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The amount arrangements would then be <math>(8!/(2!3!3!))</math>=<math>560</math>. | ||
| + | Adding up all the possibilities from cases 1),2), and 3) above would yield <math>280+420+560</math>=<math>\fbox {\textbf {(D)} 1,260}</math> as our response. | ||
==See Also== | ==See Also== | ||
Revision as of 03:28, 10 February 2019
Contents
Problem
A child builds towers using identically shaped cubes of different color. 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
Arranging eight cubes is the same as arranging the nine cubes first and then removing the last cube. Every arrangement of nine cubes corresponds to another arrangement that can be used (the first eight cubes in that order and the last cube is discarded). Thus, we get 9!. However, we overcounted because the red cubes can be permuted to have the same overall arrangement and the same with the blue and green cubes. Thus, we have to divide by the 
ways to arrange the red cubes 
ways to arrange the blue cubes, and 
ways to arrange the green cubes. Thus we have 
= 
different arrangements of towers.
- ViolinGod
Note:
This is 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
We can do this on a case-by-case basis. We divide the problem into three cases, each representing one cube to be excluded
1) The red cube is excluded. This gives us the problem of arranging one red cube, three blue cubes, and four green cubes. The amount of arrangements would then be 
. Note that we do not need to multiply by the number of red cubes because there is no way to discriminate between the first red cube and the second.
2) The blue cube is excluded. This gives us the problem of arranging two red cubes, two blue cubes, and four green cubes. The amount of arrangements would then be 
=
.
3) The green cube is excluded. This gives us the problem of arranging two red cubes, three blue cubes, and three green cubes. The amount arrangements would then be 
=
.
Adding up all the possibilities from cases 1),2), and 3) above would yield 
= as our response.
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. 
