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Difference between revisions of "2018 AMC 10A Problems/Problem 4"
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There are <math>4</math> ways to place <math>3</math> nondistinguishable classes into <math>6</math> periods such that no two classes are in consecutive periods. For each of these ways, there are <math>3! = 6</math> orderings of the classes among themselves.
 
There are <math>4</math> ways to place <math>3</math> nondistinguishable classes into <math>6</math> periods such that no two classes are in consecutive periods. For each of these ways, there are <math>3! = 6</math> orderings of the classes among themselves.
  
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Therefore, there are <math>4 \times 6 = \boxed{\mathrm{(E) \ } 24}</math> ways to choose the classes.
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Therefore, there are <math>4 \cdot 6 = \boxed{\mathrm{(E) \ } 24}</math> ways to choose the classes.
  
 
==Solution 2==
 
==Solution 2==
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<math>XOXOXO</math>
 
<math>XOXOXO</math>
  
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There are 6 ways to arrange the first class.
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There are 6 ways to place the first class.
  
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There are 4 ways to arrange the second class.
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There are 4 ways to place the second class.
  
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There is 1 way to arrange the third class.
+
There is 1 way to place the third class.
  
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We multiply <math>6*4*1= (E)  24</math>
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We multiply <math>6 \cdot 4 \cdot 1= \boxed{(E)  24}</math>
  
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-Baolan
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—Baolan
  
 
==See Also==
 
==See Also==

Revision as of 20:06, 8 February 2018

Problem

How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)

$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution 1

We must place the classes into the periods such that no two balls are in the same period or in consecutive period.

Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes, when periods cannot be consecutive:

Periods $1, 3, 5$

Periods $1, 3, 6$

Periods $1, 4, 6$

Periods $2, 4, 6$

There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.

Therefore, there are $4 \cdot 6 = \boxed{\mathrm{(E) \ } 24}$ ways to choose the classes.

Solution 2

First draw 6 $X$'s representing the 6 periods.

$XXXXXX$

Let the $O$'s represent the classes that occupy each period.

$XOXOXO$

There are 6 ways to place the first class.

There are 4 ways to place the second class.

There is 1 way to place the third class.

We multiply $6 \cdot 4 \cdot 1= \boxed{(E) 24}$

—Baolan

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

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