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Re: There are n people competing in a chess tournament. Each com [#permalink]
are these questions above 165+
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There are n people competing in a chess tournament. Each com [#permalink]
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I think it's medium one.

Remember this :

When there are \(n\) teams in a tournament and they are supposed to compete each other once, then the no of matches played will be nC2 = \(\frac{n(n-1)}{2}\)

Now in this question, we are told that the teams play \(k\) times, so the answer will be \(k\) * nC2 = \(k * \frac{n(n-1)}{2}\)

Hope this clears your doubt. Please ask if not.

Regards

void wrote:
are these questions above 165+
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Re: There are n people competing in a chess tournament. Each com [#permalink]
there are some standard formulae with that we can solve...
if there are please link it below....
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Re: There are n people competing in a chess tournament. Each com [#permalink]
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Check this out. It's a great compilation.

https://gre.myprepclub.com/forum/gre-math- ... -2609.html

void wrote:
there are some standard formulae with that we can solve...
if there are please link it below....
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Re: There are n people competing in a chess tournament. Each com [#permalink]
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