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Difference between revisions of "2013 AMC 10A Problems/Problem 24"
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*Credit to the Math Jam for this solution
 
*Credit to the Math Jam for this solution
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Let us label the players of the first team <math>A</math>, <math>B</math>, and <math>C</math>, and those of the second team, <math>X</math>, <math>Y</math>, and <math>Z</math>.
 
Let us label the players of the first team <math>A</math>, <math>B</math>, and <math>C</math>, and those of the second team, <math>X</math>, <math>Y</math>, and <math>Z</math>.
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Let us first consider how to organize A's matches, <math>AX</math>, <math>AX</math>, <math>AY</math>, <math>AY</math>, <math>AZ</math>, and <math>AZ</math>.  Because we have three duplicates, there are <math>\frac{6!}{2!*2!*2!} = 90</math> ways to organize A's matches.
 
Let us first consider how to organize A's matches, <math>AX</math>, <math>AX</math>, <math>AY</math>, <math>AY</math>, <math>AZ</math>, and <math>AZ</math>.  Because we have three duplicates, there are <math>\frac{6!}{2!*2!*2!} = 90</math> ways to organize A's matches.
  
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Now, consider <math>B</math> and <math>C</math>.  WLOG assume that A's matches were <math>XXYYZZ</math>, as we will multiply by <math>90</math> the end anyways, and that, in the first round, <math>B</math> played <math>Y</math> and C played <math>Z</math>.
+
Now, consider <math>B</math> and <math>C</math>.  WLOG assume that A's matches were <math>XXYYZZ</math>, as we will multiply by <math>90</math> the end anyways, and that, in the first round, <math>B</math> played <math>Y</math> and <math>C</math> played <math>Z</math>.
 +
 
 +
There are two cases.
 +
 
 +
1.  <math>B</math> plays <math>Y</math> again in the second round (and <math>C</math> plays <math>Z</math> in the second round)
 +
 
 +
In this case, the rest of the matches are forced, as <math>C</math> must play <math>X</math> in both of rounds <math>3</math> and <math>4</math> (as it has already played <math>Z</math> twice) and same with rounds <math>5</math> and <math>6</math> with <math>B</math> and <math>Y</math>.  Thus, there is only one option.
 +
 
 +
2.  <math>B</math> plays <math>Z</math> in the second round (and <math>C</math> plays <math>Y</math> in the second round)
 +
 
 +
In this case, <math>B</math> can play <math>Z</math> in either round <math>3</math> or <math>4</math> and <math>Y</math> in either round <math>5</math> or <math>6</math>, so there are <math>2(2) = 4</math> options.
  
 +
Thus, with <math>B</math> playing <math>Y</math> in the first round, there are <math>4+1 = 5</math> options.  Multiplying this by <math>2</math> for the case where <math>B</math> plays <math>Z</math> in the first round, we get <math>10</math> options.
  
 +
Finally, to get our final answer, we multiply <math>10 * 90 = 900</math> ways to organize the matches, <math>\textbf{(E)}</math>.
  
  

Revision as of 14:18, 8 February 2013

Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?

$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$

Solution

  • Credit to the Math Jam for this solution


Let us label the players of the first team $A$, $B$, and $C$, and those of the second team, $X$, $Y$, and $Z$.

Let us first consider how to organize A's matches, $AX$, $AX$, $AY$, $AY$, $AZ$, and $AZ$. Because we have three duplicates, there are $\frac{6!}{2!*2!*2!} = 90$ ways to organize A's matches.

Now, consider $B$ and $C$. WLOG assume that A's matches were $XXYYZZ$, as we will multiply by $90$ the end anyways, and that, in the first round, $B$ played $Y$ and $C$ played $Z$.

There are two cases.

1. $B$ plays $Y$ again in the second round (and $C$ plays $Z$ in the second round)

In this case, the rest of the matches are forced, as $C$ must play $X$ in both of rounds $3$ and $4$ (as it has already played $Z$ twice) and same with rounds $5$ and $6$ with $B$ and $Y$. Thus, there is only one option.

2. $B$ plays $Z$ in the second round (and $C$ plays $Y$ in the second round)

In this case, $B$ can play $Z$ in either round $3$ or $4$ and $Y$ in either round $5$ or $6$, so there are $2(2) = 4$ options.

Thus, with $B$ playing $Y$ in the first round, there are $4+1 = 5$ options. Multiplying this by $2$ for the case where $B$ plays $Z$ in the first round, we get $10$ options.

Finally, to get our final answer, we multiply $10 * 90 = 900$ ways to organize the matches, $\textbf{(E)}$.


See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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