Difference between revisions of "2019 AMC 10A Problems/Problem 17"
(→Solution) |
(→Solution) |
||
| Line 9: | Line 9: | ||
- ViolinGod | - ViolinGod | ||
| + | |||
| + | Note: | ||
| + | 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> | ||
==See Also== | ==See Also== | ||
Revision as of 19:25, 9 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
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)
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. 
