Suppose that each pair $I,J\!\subseteq[n]=\{1,\ldots,n\}$ for which $$\max\,([n]\!\setminus\!(I\!\cup\!J)) < \max (I\!\cap\!J) \tag{1}$$ contributes $t^{|I|+|J|}$ to a generating function, and each pair $I,J\!\subseteq[n]$ for which $$\max\,([n]\!\setminus\!(I\!\cup\!J)) > \max (I\!\cap\!J) \tag{2}$$ contributes $t^{|I|+|J|+1}$ to the same generating function. How can I show that this generating function is the one at the bottom of this calculation (taken from here):
When $I\cup J=[n]$, this is a special case (that you need not be concerned about) and only $I\!\cap\!J\!=\!\emptyset$ is allowed, so we are counting all choices of $I\!\subseteq\![n]$, of which there are $2^n$, hence we get $2^nt^{|I|+|J|}+2^nt^{|I|+|J|+1}= 2^nt^n+2^nt^{n+1} = (1+t)(2t)^n$, which is the first summand.
But when $I\cup J\subsetneq[n]$, is $L=I\cup J$ or $L=(I\!\cup\!J)\setminus(I\!\cap\!J)$? Where do $$t^3(2t)^{|L|}(1+t^2)^{n-|L|-1} \text{ and } (2t)^{|L|}(1+t^2)^{n-|L|-1} \text{ come from???}$$ I do not understand why $(1)$ should be equivalent to $(1')$ $\max([n]\!\setminus\!(I\!\cup\!J)) \notin K$. I suspect that $K\!=\!I\!\cap\!J$, and that in $(1)$ we have arbitrary $I,J$, but in $(1')$ we have $I=I\!\setminus\!I\!\cap\!J$ and $J=J\!\setminus\!I\!\cap\!J$.