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Difference between revisions of "2019 AMC 10A Problems/Problem 17"

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<math>\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320</math>
 
<math>\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320</math>
 +
 +
== Video Solution ==
 +
https://youtu.be/3MiGotKnC_U?t=1083
 +
~ ThePuzzlr
 +
 +
== Video Solution ==
 +
https://youtu.be/0W3VmFp55cM?t=612
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution ==
 +
https://youtu.be/Sk0Gm__kLpo
 +
 +
Education, the Study of Everything
  
 
==Solution 1==
 
==Solution 1==
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''Note'':
 
''Note'':
−
This can be written more compactly as
+
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>
  
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We can divide the problem into three cases, each representing one cube to be excluded:
 
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 <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.
+
'''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 <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.
  
 
'''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>.
 
'''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>.
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Adding up the individual cases from above gives the answer as <math>280+420+560=\boxed{\textbf{(D) } 1,260}</math>.
 
Adding up the individual cases from above gives the answer as <math>280+420+560=\boxed{\textbf{(D) } 1,260}</math>.
−
 
−
==Solution 3 (guessing)==
 
−
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 10:11, 19 October 2021

Problem

A child builds towers using identically shaped cubes of different color. 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.)

$\textbf{(A) } 24 \qquad\textbf{(B) } 288 \qquad\textbf{(C) } 312 \qquad\textbf{(D) } 1,260 \qquad\textbf{(E) } 40,320$

Video Solution

https://youtu.be/3MiGotKnC_U?t=1083 ~ ThePuzzlr

Video Solution

https://youtu.be/0W3VmFp55cM?t=612

~ pi_is_3.14

Video Solution

https://youtu.be/Sk0Gm__kLpo

Education, the Study of Everything

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 $9!$. 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 $2!$ ways to arrange the red cubes, the $3!$ ways to arrange the blue cubes, and the $4!$ ways to arrange the green cubes. Thus we have $\frac {9!} {2! \cdot 3! \cdot 4!} = \boxed{\textbf{(D) } 1,260}$ different possible towers.

Note: this can be written more compactly as \[\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}\]

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 of possible arrangements is $\frac{8!}{4!\cdot3!}=280$. 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 $\frac{8!}{2!\cdot2!\cdot4!}=420$.

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 $\frac{8!}{2!\cdot3!\cdot3!}=560$.

Adding up the individual cases from above gives the answer as $280+420+560=\boxed{\textbf{(D) } 1,260}$.

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

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