Question

In a chess tournament, each of six players will play every other player exactly once. How many players will be played during the tournament?
a)36
b)30
c)15
d)12

Answer 1

Hint:In this question, we have to find the number of matches such that each of six players will play every other player exactly once. Therefore, as for a match, we need to choose two players and as the ordering in which the players are chosen for a particular match, the answer should be equal to the combination of 6 objects taken 2 at a time. Therefore, we can use the formula for combination to get the required answer to this question.

Complete step-by-step answer:
In this question, we have to find the total number of chess matches. As for a particular match, two players are required, we need to find the number of ways two players can be chosen. However, in a particular pair, the ordering is not important, for example, the match between first and second player is the same as the match between second and first player. Thus, the total number of matches should be equal to the total number of combinations of 6 players taken 2 at a time. Therefore,
Total number of matches= $ ^{6}{{C}_{2}}...................................(1.1) $
Where $ ^{n}{{C}_{r}} $ represents the number of combinations of n objects taken r at a time and is given by
 $ ^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}...........................(1.2) $
And for a number n, $ n!=n\times \left( n-1 \right)\times (n-2)\times ...\times 1 $ ……………………(1.3)
Therefore, using equation (1.2) in equation (1.1), we obtain
Total number of matches= $ ^{6}{{C}_{2}}=\dfrac{6!}{2!\left( 6-2 \right)!}=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{\left( 2\times 1 \right)\times \left( 4\times 3\times 2\times 1 \right)}=\dfrac{6\times 5}{2}=15 $
Which matches option (c) of the question. Thus, option (c) is the correct answer.

Note: In this question, we took combination as the order of choosing two players for a particular match was not important. However, in cases where the order is important, such as choosing the ways in which the winner and runners up can be decided, we should use permutation of the six objects taken two at a time $ \left( ^{6}{{P}_{2}}=\dfrac{6!}{\left( 6-2 \right)!} \right) $ rather than combination.