2013 AMC 10A Problem 24

Question

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?

540

600

720

810

900

Answer: E

Concepts:arrangements with restrictions multiset permutations casework

Difficulty rating: 2390

Small hint:

Fix one player’s six-round schedule first

Big hint:

For the second player, count strings avoiding the first player’s opponent in every round

Solution:

Label Central’s players

A,B,C

and Northern’s players

X,Y,Z

.

Player

A

’s six-round opponent string contains two each of

X,Y,Z

, so it can be chosen in

\frac{6!}{2!2!2!}=90

ways.

For a fixed

A

-string, say

XXYYZZ

, player

B

’s string must also contain two each of

X,Y,Z

and cannot match

A

’s opponent in any position.

If

B

’s first two entries are

Y,Z

in either order, then the middle two entries must be

X,Z

in either order and the last two must be

X,Y

in either order, giving

2^3=8

strings. The two remaining possibilities are

YYZZXX

and

ZZXXYY

, for

10

total

B

-strings.

Once

A

and

B

are scheduled,

C

’s schedule is forced. Hence there are

90\cdot10=900

schedules.

Thus, E is the correct answer.