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

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(Solution)
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- ViolinGod
 
- ViolinGod
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Note:
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This is written more compactly as
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<cmath>\binom{9}{2,3,4}=\binom{9}{2}\binom{9-2}{3}\binom{9-(2+3)}{4} = \boxed{1,260}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 18:25, 9 February 2019

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.)

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

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

- ViolinGod

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

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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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