Need some help with this.

How do we solve the following problem?

There are $n$ players participating in a sports tournament. Each player plays with every other player and each game ends in a win for one of the players (no draws). Prove that the players can be arranged in the following sequence: $P_1, P_2, P_3, \ldots , P_n$ where, player $P_i$ has beaten the player $P_{i+1} (i=1,2,3, \ldots , n-1)$ in at least one game.

#Combinatorics

Easy Math Editor

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Comments

I think, I have got a reasonable proof.I will be using induction.Consider the base case of two players.In two players, there will be a winner and a loser, so we can definitely form the sequence.Now,assume that we can form the given sequence for k players.So, we have the sequence P1,P2,P3,.........,Pk.We must now show that it is always possible to insert P (k+1) somewhere between the gaps of Pi (i=1,2,3.......,k).We know that P (k+1) will play exactly k games with each of P1,P2,......Pk.So,there will be k outcomes,where each of outcome is win (W) or lose (L).So,we consider the string of k letters where each letter is L or W.If at,say r(th) position,there is W,then it means that P (k+1) defeated Pr.Now,if first letter of the string is W,then P (k+1) will precede P1.Now,consider that the first letter of the string is L.If second letter is W,then P (k+1) comes between P1 and P2.Suppose second letter is L.Now,if third letter is W, then P (k+1) comes between P2 and P3.This logic can be extended,so as to get that,we can always find a consecutive L and W, if not, every letter of the string will be L.In that case P (k+1) will succeed Pk.So,it has been shown that it is always possible to form a sequence with (k+1) players.So,the question has been proved using induction.Does this solution seem fine?

Indraneel Mukhopadhyaya - 5 years, 1 month ago

Hmm... This seems fine. It resembles the proof I'd written in the exam very much (but I missed one crucial point in my written proof 😣).

Btw, is there partial credit?

A Former Brilliant Member - 5 years, 1 month ago

Yes,definitely there is partial credit. Does my solution take into account,the crucial point you are talking about?

Indraneel Mukhopadhyaya - 5 years, 1 month ago

@Indraneel Mukhopadhyaya – Yeah, it does. Only after you gave your proof did I realise that I was pretty close to the actual proof.

Once again, thanks for the proof!

A Former Brilliant Member - 5 years, 1 month ago

@Indraneel Mukhopadhyaya

Did you write the CMI exam?

A Former Brilliant Member - 5 years ago

Hey, @Indraneel Mukhopadhyaya when is your interview for ISI?

A Former Brilliant Member - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$