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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
1 year, 11 months ago

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Are each of the $$x_i$$ distinct? Because then you have one solution , the identity function. · 1 year, 11 months 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 ... · 1 year, 11 months ago

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. · 1 year, 11 months ago

Great! That's an invariance argument. Staff · 1 year, 11 months ago

if $$n$$ is odd then how is $$n=2 ?$$ , in the set A? · 1 year, 11 months 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 .. · 1 year, 11 months ago

ok ..i am sorry.....i thought that x(n) is a general term.....did not look at the end of the set. :( · 1 year, 11 months ago

it's okay bro .. please try it now.. and help me · 1 year, 11 months ago

yeah.. · 1 year, 11 months ago