May not be a best answer but here is an attempt

Consider a set A with 'n' elements, A = {a1,a2.....,an}

Consider a set B with 'n+1' elements, B = A + {x} = {a1,a2.....,an,x} - Here 'x' will have a special meaning for us.

Now you RHS = n^2 is simply the number of ways you can arrange 'n' elements from set A in 2 seats. repititions are allowed.

Now there is another way to count the same.

Select two elements from set B. i.e. (n+1)C2 ways.

Here in a selection, if we get element 'x' in the selection we interpret is as a repeat. So, for us {x,a1} gets mapped to {a1,a1}

But here we have missed a few cases for e.g. we counted {a1,a2} but did not count {a2,a1} [as we are using only selections]. So we need to add that

Now note the order will not matter when there is a repeatition. So excluding element 'x' there are nC2 selection. Which need to added to get the final answer.

So we get

(n+1)C2 + nC2 = n^2

It is easy to check that any pair like (a,b) is counted same number of times in LHS and RHS.

if a = b;

RHS count = 1

Now this pair can't come in the second selection, so it is counted only once (as {a,x} order doesn' matter)

similarly you can show if a<>b each is counted twice.

Hope this helps - maybe someone can show a more elegant way to put it though.