suppose x,y simultaneously cross z. from z's point is it a single beyond experiance or double beyond experiance? also "everyone just had two "beyond" experience" Do you mean exactly two beyond experiance or atmost two beyond experiance?

sumanguha wrote: suppose x,y simultaneously cross z. from z's point is it a single beyond experiance or double beyond experiance? Double.one cross makes two "beyond" experiance Exactly two beyond experiance

I think the problem is incorrect or i don't understand it correctly!! because see the following WLOG rank the velocities as following v(1)=<v(2)=<....=<v(7) so see v(1) can't cross anybody it can only be crossed. so only two among others say i,j (WLOG) with velocity v(i), v(j) can cross him so v(i),v(j)>=v(1) and all other v(k)=v(1) but v(i),v(j)>v(1) strict inequality holds as v(1) have to have exactly two beyond experiance. but then v(i) have to have atleast five beyond experiance as he crosses atleast five guys. so it violates the given two beyond experiance condition.

sumanguha wrote: I think the problem is incorrect or i don't understand it correctly!! because see the following WLOG rank the velocities as following v(1)=<v(2)=<....=<v(7) so see v(1) can't cross anybody it can only be crossed. so only two among others say i,j (WLOG) with velocity v(i), v(j) can cross him so v(i),v(j)>=v(1) and all other v(k)=v(1) but v(i),v(j)>v(1) strict inequality holds as v(1) have to have exactly two beyond experiance. but then v(i) have to have atleast five beyond experiance as he crosses atleast five guys. so it violates the given two beyond experiance condition. WLOG rank the velocities as following v(1)=<v(2)=<....=<v(7) this is wrong,the numbers can only represent starting sequence and their speeds are unrelated

I assume that the formulation goes like this 1 start first then 2 then 3......finally 7 Is my understanding correct ? then see then suppose the following is the ending sequence 3<4<1<2<7<6<5 ie 3 reaches at first then 4 ...like this . then everyone has exactly two beyond experiance but there are seven different rankings!!

sumanguha wrote: I assume that the formulation goes like this 1 start first then 2 then 3......finally 7 Is my understanding correct ? then see then suppose the following is the ending sequence 3<4<1<2<7<6<5 ie 3 reaches at first then 4 ...like this . then everyone has exactly two beyond experiance but there are seven different rankings!! I proved only two: \[\begin{gathered} \left\{ {7,6,5,4,3,2,1} \right\} \to \left\{ {5,4,7,6,1,2,3} \right\} \hfill \\ \left\{ {7,6,5,4,3,2,1} \right\} \to \left\{ {5,6,7,2,1,4,3} \right\} \hfill \\ \end{gathered} \]

i am sorry. there is a huge misunderstanding from my side. you mean two different set of rankings i munderstood within a set of ranking only two different rankings for seven skiers.

Seven skiers with numbers 1,2, . . . ,7 left the start one by one and walked the distance - each at their own constant speed. It turned out that each skier participated in overtaking exactly twice. (In each overtaking, exactly two skiers participate - the one who overtakes and the one who is overtaken.) At the end of the race, a ranking must be drawn up consisting of numbers skiers in finishing order. Prove that you are in the race with the described properties, no more than two different ranking can be obtained.

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