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11 votes
11 votes

Ten teams participate in a tournament. Every team plays each of the other teams twice. The total number of matches to be played is  

  1. $20$
  2. $45$
  3. $60$
  4. $90$
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4 Answers

Best answer
12 votes
12 votes
We have $10$ teams.

First team can play with any other team in $9$ ways.
Similarly second can play with any other in $8$ ways (excluding the first team), $3^{rd}$ team can play $7$ games and so on.

So, total number of games possible $=9+8+7\ldots +1=\dfrac{n(n+1)}{2} =\dfrac{10\times 9}{2}=45$

Since, each team plays $2$ games against each other, total games $=45\times 2=90$

Correct Answer: $D$
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9 votes
9 votes
2 teams can be selected in $\binom{10}{2}$ ways... each team will play 2 match with other teams... so $2*\binom{10}{2}$ = 90 ... correct me if i am wrong ..
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2 votes
2 votes
2 teams can be selected in 10C2 ways.But this is only 1 match between them.

Now each of these pairs will play against each other twice . So 10C2 *2 = 90
1 votes
1 votes
10 teams are there, each team played total 18 matches( two matches with each remaining 9 teams , so total 2*9=18).

So, total matches = (10*18)/2 [ dividing by two because 10 *18 will count each matches twice, from teamX to teamY once and from teamY to teamX one more time] = 90 (Answer D)
Answer:

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