Question

There is a chess team with four players {A, B, C, D} training for a competition. The coach wants all players to play each other. Ignore which player is chosen to take the first move.

1. Do we use a counting rule for combinations or for permutations to calculate the total number of possible games if all players must play each other once? (Only one is correct.)

2. Calculate by hand (you can use a calculator but show some work), how many games will there be if all players must play each other once?

3. If we have players {A, B, C, D} and two players will be selected at random for a challenge. What is the probability that players A and B are chosen to play? This follows from your answer from part (b). (By the way, this is the same probability of any two players being chosen.)

Answer 1

Solution 1

We use a counting rule for combinations to calculate total number of possible games. This is because we ignore which player makes the first move and also all players must play each other once.

(In combinations, what matters is the elements chosen and the order in which they are chosen is immaterial. For example, AB and BA would mean the same thing under combiantions

whereas

In permutations, the order in which the elements are chosen is also important. For example, AB would be considered different from BA under permutations.)

Solution 2

We list all the possible combinations as follows,

First we observe the games played by A : AB, AC and AD

Now, we observe the games played by B(except for with A, since it has already been counted) : BC and BD

Finally, one game will be played between C and D : CD

Thus, all the possible games are : { AB, AC, AD, BC, BD, CD}, i.e., a total of 6 games will be played if all the players must play each other once.

We could calculate the no. of games by using the combinations formula (no. of ways in which 2 players can be chosen out of 4 players) of :

Solution 3

Since, there are 6 possible combinations and each combination of players is equally likely. Thus:

P(A and B are chosen to play) = P(AB) = (No. of favourable outcomes)/(Total outcomes) = 1/6

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