Hi, there are several solutions.
I prefer this:
after your question about ages (Q0)
create a question of type equation (eqQ0) with this:
{substr( join(if(Q0_SQ001.NAOK>7,"A",""),if(Q0_SQ002.NAOK>7,"B",""),if(Q0_SQ003.NAOK>7,"C",""),if(Q0_SQ004.NAOK>7,"D",""),if(Q0_SQ005.NAOK>7,"E","")),rand(1,countifop(">",7,that.Q0.NAOK))-1,1)}
which will give you a letter "A" - child 1, "B" - child 2,...
If you provide less then 10 options you may use digits instead.
Now what does this equation do?
with the join function you build a string that contains only the children >7
join(if(Q0_SQ001.NAOK>7,"A",""),if(Q0_SQ002.NAOK>7,"B",""),if(Q0_SQ003.NAOK>7,"C",""),if(Q0_SQ004.NAOK>7,"D",""),if(Q0_SQ005.NAOK>7,"E",""))
you create a random number between 1 and the number of children>1
rand(1,numChildren)
The number of children you get by the countifop function ("<",7,Q0), count Q0, if <7
At last you extract the letter from your string that refers to your random number.
Remember: substr starts counting with 0.
Afterwards there is your group with the additional questons.
Here you can display the number of the child by micro-tayloring.
"Now we talk about child {if(eqQ0=="A","Child 1",if(eqQ0=="B","Child 2",if(eqQ0=="C","Child 3",if(eqQ0=="D","Child 4","Child 5"))))}
If you have you can enter the name of the child or whatever.
Adapt it to your needs. I only showed for 5 answer options (children)
So you will get a very short survey. Only one group for your additional questions.
And in the dataset you'll habe the letter to distinguish between the children.
And additionally you have to check if there are children>7
Joffm