Understand the problem

Suppose that in a sports tournament featuring n players, each pair
plays one game and there is always a winner and a loser (no draws).
Show that the players can be arranged in an order P1, P2, . . . , Pn such
that player Pi has beaten Pi+1 for all i = 1, 2, . . . , n.  

Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016. Subjective Problem no.1.


Difficulty Level

7 out of 10

Suggested Book

Start with hints

Do you really need a hint? Try it first!

we define a sequence \(p_1\)>\(p_2\)>\(p_3\)………..>\(p_n\) . such that \(p_i>p_{i+1}\) implies \(p_i\) has beaten \(p_{i+1}\)  

let assume that the statement is true for n=k  i.e \(p_1> p_2>p_3………p_k\) now look into the case of \(p_{k+1}\)    

if \(p_{k+1}\) > \(p_1\) or \(p_k\)>\(p_{k+1}\) we are done (why?)

 if \(p_k\)<\(p_{k+1}\) or \(p_{k+1}\)<\(p_1\) then  we can find out at least one pair \(p_i\) >\(p_{i+1}\) such that \(p_i\)> \(p_{k+1}\) >\(p_{i+1}\) , we can just find it by shifting \(p_{k+1}\) term by term [ you can make some small experiment ]. other wise it contradict our initial assumption that \(p_k\)<\(p_{k+1}\) or \(p_{k+1}\)<\(p_1\) [ why?]   hence we are done 

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