Difference between revisions of "2019 AMC 10A Problems/Problem 17"
Sevenoptimus (talk | contribs) m (Further fixed formatting) |
Sevenoptimus (talk | contribs) m (Fixed grammar and formatting) |
||
| Line 6: | Line 6: | ||
==Solution 1== | ==Solution 1== | ||
| − | Arranging eight cubes is the same as arranging the nine cubes first and then removing the last cube. | + | 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 <math>9!</math>. 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 <math>2!</math> ways to arrange the red cubes, the <math>3!</math> ways to arrange the blue cubes, and the <math>4!</math> ways to arrange the green cubes. Thus we have <math>\frac {9!} {2! \cdot 3! \cdot 4!} = \boxed{\textbf{(D) } 1,260}</math> different possible towers. |
''Note'': | ''Note'': | ||
| − | This | + | This can be 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== | ==Solution 2== | ||
| − | We can | + | We can divide the problem into three cases, each representing one cube to be excluded: |
| − | 1 | + | '''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 opossible arrangements is <math>\frac{8!}{4!\cdot3!}=280</math>. 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. |
| − | 2 | + | '''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 <math>\frac{8!}{2!\cdot2!\cdot4!}=420</math>. |
| − | 3 | + | '''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 <math>\frac{8!}{2!\cdot3!\cdot3!}=560</math>. |
| − | Adding up | + | Adding up the individual cases from above gives the answer as <math>280+420+560=\boxed{\textbf{(D) } 1,260}</math>. |
==Solution 3 (guessing)== | ==Solution 3 (guessing)== | ||
| − | If you're running out of time, notice that <math>A</math>, <math>B</math>, and <math>C</math> are way too small, and <math>E</math> would make no sense since it would simply be <math>8!</math>, as if there were no restrictions. Thus the answer | + | If you're running out of time, notice that choices <math>A</math>, <math>B</math>, and <math>C</math> are way too small, and choice <math>E</math> would make no sense since it would simply be <math>8!</math>, as if there were no restrictions. Thus, by educated guessing and elimination, the correct answer must be <math>\boxed{\textbf{(D) } 1,260}</math>. |
| + | |||
| + | ''Note'': this strategy is not recommended! | ||
==See Also== | ==See Also== | ||
Revision as of 00:16, 27 February 2019
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 1
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
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 opossible 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 (guessing)
If you're running out of time, notice that choices 
, 
, and 
are way too small, and choice 
would make no sense since it would simply be 
, as if there were no restrictions. Thus, by educated guessing and elimination, the correct answer must be 
.
Note: this strategy is not recommended!
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. 
