Difference between revisions of "2013 AMC 10A Problems/Problem 24"

(Solution)
Line 5: Line 5:
 
==Solution==
 
==Solution==
  
*Credit to the Math Jam for this solution
+
*Credit saurabh_g for this solution
  
  
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>. One way of scheduling all six rounds could be:
  
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.
+
Round1---> AX BY CZ
 +
Round2---->AX BZ CY
 +
Round3---->AY BX CZ
 +
Round4---->AY BZ CX
 +
Round5---->AZ BX CY
 +
Round6---->AZ BY CX
  
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>.
+
The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permuting those 6 rounds and that can be done in <math>6!</math>=<math>720</math> ways.
  
There are two cases.
+
So the answer is <math>720</math>
 
 
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 = \boxed{\textbf{(E) }900}</math> ways to organize the matches.
 
  
 
==See Also==
 
==See Also==

Revision as of 14:10, 20 January 2015

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 saurabh_g 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 $Z$. One way of scheduling all six rounds could be:

Round1---> AX BY CZ
Round2---->AX BZ CY
Round3---->AY BX CZ
Round4---->AY BZ CX
Round5---->AZ BX CY
Round6---->AZ BY CX

The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permuting those 6 rounds and that can be done in $6!$=$720$ ways.

So the answer is $720$

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS