Question

There are 8 tennis players in a round robin tournament, which means each team will play every other team once. How many matches will be held during the 8-person round robin tennis tournament?​

Answer 1

Answer:  28

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Explanation:

Method 1

Imagine a table with 8 rows and 8 columns to represent all possible match-ups. You can actually draw out this table or just think of it as a thought experiment.

There are 8*8 = 64 entries in the table. Along the northwest diagonal, we have each team pair up with itself. This is of course silly and impossible. We cross off this entire diagonal so we drop to 64-8 = 56 entries.

Then notice that the lower left corner is a mirror copy of the upper right corner. A match-up like AB is the same as BA. So we must divide by 2 to get 56/2 = 28 different matches.

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Method 2

There are 8 selections for the first slot, and 8-1 = 7 selections for the second slot. We have 8*7 = 56 permutations and 56/2 = 28 combinations.

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Method 3

Use the nCr combination formula with n = 8 and r = 2

$n C r = \frac{n!}{r!(n-r)!}\\\\8 C 2 = \frac{8!}{2!*(8-2)!}\\\\8 C 2 = \frac{8!}{2!*6!}\\\\8 C 2 = \frac{8*7*6!}{2!*6!}\\\\ 8 C 2 = \frac{8*7}{2!}\\\\ 8 C 2 = \frac{8*7}{2*1}\\\\ 8 C 2 = \frac{56}{2}\\\\ 8 C 2 = 28$

There are 28 combinations possible. Order doesn't matter (eg: match-up AB is the same as match-up BA).

Notice how the (8*7)/2 expression is part of the steps shown above in the nCr formula.