Difference between revisions of "2007 AMC 10B Problems/Problem 20"
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Finally, the last square has <math>9</math> remaining choices. | Finally, the last square has <math>9</math> remaining choices. | ||
| − | The number of ways to choose <math>3</math> squares is <math>25 \cdot 16 \cdot 9,</math> but the order in which you chose the squares does not matter as the blocks are | + | The number of ways to choose <math>3</math> squares is <math>25 \cdot 16 \cdot 9,</math> but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by <math>3!</math>. |
<cmath>\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 600}</cmath> | <cmath>\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 600}</cmath> | ||
Revision as of 13:37, 4 June 2021
Contents
Problem
A set of 
square blocks is arranged into a 
square. How many different combinations of 
blocks can be selected from that set so that no two are in the same row or column?
Solution 1
There are ways to choose the first square. The four remaining squares in its row and column and the square you chose exclude nine squares from being chosen next time.
There are 
remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has 
remaining choices.
The number of ways to choose squares is 
but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by 
.
![\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{\mathrm{(C) \ } 600}\]](http://latex.artofproblemsolving.com/c/c/1/cc16a325891282648c5127af09cca9adef42b279.png)
Solution 2
Once we choose our three squares, we will have occupied three separate columns 
and three separate rows. There are 
ways to choose these rows and columns.
There are ways to assign the square in column 
to a row, 
ways to assign the square in column 
to one of the remaining two rows, and poor square in column 
doesn't get to choose. 
In total, we have ![\[{5 \choose 3} \times {5 \choose 3} \times 3!\]](http://latex.artofproblemsolving.com/b/c/2/bc24d0f0e0a7dc502555585ac4f647af4acdecdb.png)
which totals out to 
.
![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
~HappyHuman
Solution 3 (Answer choices)
We know that there are 
ways to choose three blocks. However, the restriction clearly limits the number of ways we can choose our blocks. Hence, only 
, 
, or 
are the correct choices. Clearly, there are more than 
ways, thus yielding 
ways.
Video Solution 1
https://youtu.be/0W3VmFp55cM?t=4921
~ pi_is_3.14
Video Solution 2
https://youtu.be/5UojVH4Cqqs?t=3460
~ pi_is_3.14
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
