A question on Permutation and Combination.

[cooldudeachyut]

Question - In a tennis tournament in which every pair has to play with every other pair, 10 players are playing. Find the number of games played.

My attempt - The total number of possible pairs will be ¹⁰C₂ = 45. Total number of pairs fixing 1 team = ⁸C₂ = 28. So total number of matches should be 45x28 = 1260 but the answer in my textbook is 630.

 

[Ishuda]

Question - In a tennis tournament in which every pair has to play with every other pair, 10 players are playing. Find the number of games played.

My attempt - The total number of possible pairs will be ¹⁰C₂ = 45. Total number of pairs fixing 1 team = ⁸C₂ = 28. So total number of matches should be 45x28 = 1260 but the answer in my textbook is 630.

Yes, their are ¹⁰C₂ = 45 possible pairs and p₁ (player 1) plays with all pairs in which he isn't a member. So with p₂ as a partner he plays ⁸C₂ games, with p₃ as a partner, he plays with ⁸C₂ pairs, ... for a total of 9 * ⁸C₂ games. Now for p₂, since we don't double count, we don't count the number of games p₁ played in and we have 8 * ⁷C₂ for p₂, 7 * ⁶C₂ for p₃, 6 * ⁵C₂ for p₄, 5 * ⁴C₂ for p₅, 4 * ³C₂ for p₆, 3 * ²C₂ for p₇. There are now 3 players left so a pairs game can't be made. So add them up
N = \(\displaystyle \Sigma_{j=2}^{j=8}\, (j+1)\,\, ^jC_2\)

 

[cooldudeachyut]

Yes, their are ¹⁰C₂ = 45 possible pairs and p₁ (player 1) plays with all pairs in which he isn't a member. So with p₂ as a partner he plays ⁸C₂ games, with p₃ as a partner, he plays with ⁸C₂ pairs, ... for a total of 9 * ⁸C₂ games. Now for p₂, since we don't double count, we don't count the number of games p₁ played in and we have 8 * ⁷C₂ for p₂, 7 * ⁶C₂ for p₃, 6 * ⁵C₂ for p₄, 5 * ⁴C₂ for p₅, 4 * ³C₂ for p₆, 3 * ²C₂ for p₇. There are now 3 players left so a pairs game can't be made. So add them up
N = \(\displaystyle \Sigma_{j=2}^{j=8}\, (j+1)\,\, ^jC_2\)

Thanks.