A way to sort elements of a set.

[Zan]

Choose n numbers from the set \[\{1;2;...;2n\}\] and sort them in that way: \[a_1<a_2<...<a_n\] Sort the other n numbers so that \[b_1>b_2>...>b_n\]
Prove \[|a_1-b_1|+|a_2-b_2|+...+|a_n-b_n|=n^2\]

[sourav das]

We claim:
$\sum_{i=1}^n |a_i-b_i|=2n+(2n-1)+....+(n+1)-n-(n-1)-(n-2)....-1=n^2$
Let $|a_i-b_i|=a_i-b_i$ for some $i$. Then $a_i>b_i>...>b_n$ and also $a_i>a_{i-1}>a_{i-2}>....>a_1$ it implies $a_i$ is greater than $(n-i+1)+(i-1)=n$ different elements from the set. It implies $a_i>n$. Thus we can prove that for $|a_i-b_i|=b_i-a_i$ that $b_i>n$ as then $b_i>a_i>a_{i-1}>...>a_1$ and $b_i>b_{i+1}>...>b_n$ same argument as before. It implies $max(a_i,b_i)>n$ hence $\sum_{i=1}^n |a_i-b_i|=(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

done.
Cool problem Zan. Wecome to BDMO forum.

[Zan]

Thank you for your reply. But could you explain the last conclusion, please?
My teacher is very demanding. I've got many other unsolved problems.

[*Mahi*]

It implies $max(a_i,b_i)>n$ hence $\sum_{i=1}^n |a_i-b_i|=(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

For each $(a_i,b_i)$ exactly one of them is greater than $n$, and thus we can say that, all of the numbers greater than $n$ has "$+$" sign, and all the numbers from $1$ to $n$ has "$-$" sign and thus the sum $\sum_{i=1}^n |a_i-b_i|$ has the value $(\sum_{i=n+1}^{2n} i)-(\sum_{j=1}^n j)=n^2$

[Zan]

Thanks a lot!!!!