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

One-onefunction(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

## Comments

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TopNewestAre each of the \( x_i \) distinct? Because then you have one solution , the identity function. – Siddhartha Srivastava · 2 years, 1 month ago

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– Karan Shekhawat · 2 years, 1 month ago

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 ...Log in to reply

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, 1 month ago

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invariance argument. – Calvin Lin Staff · 2 years, 1 month ago

Great! That's anLog in to reply

if \(n\) is odd then how is \(n=2 ?\) , in the set A? – Mahimn Bhatt · 2 years, 1 month ago

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– Karan Shekhawat · 2 years, 1 month ago

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 ..Log in to reply

– Mahimn Bhatt · 2 years, 1 month ago

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

– Karan Shekhawat · 2 years, 1 month ago

it's okay bro .. please try it now.. and help meLog in to reply

– Mahimn Bhatt · 2 years, 1 month ago

yeah..Log in to reply

@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, 1 month ago

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