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Math Challenge

Let an Set \(\displaystyle{A=\left\{ { x }_{ 1 },{ x }_{ 2 },........,{ x }_{ n } \right\} \\ n=2k+1,\quad k\in { I }^{ + }}\) ,

One-one function(s) \[\displaystyle{f:A\rightarrow A\quad \\ \left| f\left( { x }_{ 1 } \right) -{ x }_{ 1 } \right| =\left| f\left( { x }_{ 2 } \right) -{ x }_{ 2 } \right| =........=\left| f\left( { x }_{ n } \right) -{ x }_{ n } \right| }\] , then Find number of all such one-one functions ?

Help Me!

Thanks..!

Karan

Note by Karan Shekhawat
2 years, 7 months ago

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Are each of the \( x_i \) distinct? Because then you have one solution , the identity function.

Siddhartha Srivastava - 2 years, 7 months ago

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Yes It is... nothing is mention , and read question properly ... We have to Count all such possible cases , obviously f(x)=x is one among such.. But we have to find all ...

Karan Shekhawat - 2 years, 7 months ago

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OK. I only see 1 solution though. I even have a proof. Do you have a counter example?

Let \( |f(x_i) - x_i| = m \). If \( m =0 \), we have the identity function. So let us assume \( m \not = 0 \) and a function other than the identity function exists.

Then \( f(x_i) - x_i = \pm m \quad \forall i \in \{ 1,2,...,n \} \).

Divide \( A \) into \( P \) and \( Q \), where if \( f(x_i) - x_i = m \implies x_i \in P \) else \( x_i \in Q \). Since \( P \) and \( Q \) are disjoint, \( |P| + |Q| = |A| = odd \).

Now, \( \sum f(x_i) - x_i = |P|m + |Q|(-m) = (|P| - |Q|)m \)

Now, since \( |P| + |Q| \) is odd, \( |P| - |Q| \) is also odd, hence not zero. Therefore, \( (|P| - |Q|)m \) is not zero and \( \sum f(x_i) - x_i \) is not zero.

But, \( f(x) \) is clearly a permutation function, which means that \( \sum f(x_i) - x_i = 0 \), contradiction.

Therefore there exists no other function other than the identity function.

Siddhartha Srivastava - 2 years, 7 months ago

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@Siddhartha Srivastava Great! That's an invariance argument.

Calvin Lin Staff - 2 years, 7 months ago

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if \(n\) is odd then how is \(n=2 ?\) , in the set A?

Mahimn Bhatt - 2 years, 7 months ago

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so what ... ? Please read it properly bro .. n is odd Means total number of terms are odd , It dosn't mean that all terms are odd.... so be careful ..

Karan Shekhawat - 2 years, 7 months ago

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ok ..i am sorry.....i thought that x(n) is a general term.....did not look at the end of the set. :(

Mahimn Bhatt - 2 years, 7 months ago

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@Mahimn Bhatt it's okay bro .. please try it now.. and help me

Karan Shekhawat - 2 years, 7 months ago

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@Karan Shekhawat yeah..

Mahimn Bhatt - 2 years, 7 months ago

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@Shashwat Shukla @Ronak Agarwal @Raghav Vaidyanathan @Mvs Saketh @Sudeep Salgia @Calvin Linsir @Brian Charlesworth sir @Krishna Sharma @Tanishq Varshney Please help me in this if you can... any help will be appreciated ! Thanks!

Karan Shekhawat - 2 years, 7 months ago

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