Difference between revisions of "2013 AMC 10A Problems/Problem 24"
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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>. One way of scheduling all six rounds could be: | + | 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>. |
+ | i)One way of scheduling all six distinct rounds could be: | ||
Round1---> <math>AX</math> <math>BY</math> <math>CZ</math> | Round1---> <math>AX</math> <math>BY</math> <math>CZ</math> | ||
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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. | 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. | ||
− | So the | + | (ii) One can also make the schedule in such a way that two rounds are repeated |
+ | (a) | ||
+ | Round1---> <math>AX</math> <math>BZ</math> <math>CY</math> | ||
+ | Round2----><math>AX</math> <math>BZ</math> <math>CY</math> | ||
+ | Round3----><math>AY</math> <math>BX</math> <math>CZ</math> | ||
+ | Round4----><math>AY</math> <math>BX</math> <math>CZ</math> | ||
+ | Round5----><math>AZ</math> <math>BX</math> <math>CY</math> | ||
+ | Round6----><math>AZ</math> <math>BX</math> <math>CY</math> | ||
+ | |||
+ | (b) | ||
+ | Round1---> <math>AX</math> <math>BY</math> <math>CZ</math> | ||
+ | Round2----><math>AX</math> <math>BY</math> <math>CZ</math> | ||
+ | Round3----><math>AY</math> <math>BZ</math> <math>CX</math> | ||
+ | Round4----><math>AY</math> <math>BZ</math> <math>CX</math> | ||
+ | Round5----><math>AZ</math> <math>BY</math> <math>CX</math> | ||
+ | Round6----><math>AZ</math> <math>BY</math> <math>CX</math> | ||
+ | |||
+ | As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are | ||
+ | <math>6!/(2!2!2!</math> = <math>90</math> | ||
+ | |||
+ | |||
+ | So the total number of schedules = <math>720+90+90</math> =900 (Answer E) | ||
==See Also== | ==See Also== |
Revision as of 17:21, 21 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?
Solution
- Credit saurabh_g for this solution
Let us label the players of the first team , , and , and those of the second team, , , and .
i)One way of scheduling all six distinct rounds could be:
Round1---> Round2----> Round3----> Round4----> Round5----> Round6---->
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 = ways.
(ii) One can also make the schedule in such a way that two rounds are repeated (a)
Round1---> Round2----> Round3----> Round4----> Round5----> Round6---->
(b)
Round1---> Round2----> Round3----> Round4----> Round5----> Round6---->
As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are =
So the total number of schedules = =900 (Answer 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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