I am just looking through a book and I came across this question. I am hoping someone can help me clarify whether the number of arrangements would be $20 \choose 2$ or would it be some other answer.
QUESTION: A family of ten boys decides to make an alliance and marry off their boys to another family of ten girls each of whom is equal in age to one of the boys. How many arrangements are possible? How many if 6 girls chose to marry someone not their own age?
$20 \choose 2$ counts the number of ways to choose $2$ objects from $20$. In this case, it would be the number of pairs possible if you chose two children from the $20$ boys and girls, which is not what you want.
Instead, consider this: arrange the boys in line based on age (assuming they all have different ages, since there could be twins). If you arrange the girls in a line, that presents a marriage arrangement- the first girl goes with the first boy, the second girl goes with the second boy, etc. Also, each arrangement can be represented like this. So the number of arrangements is the number of ways to arrange the girls in line, which you should know.
If $6$ girls don't want to marry a boy of her age, we use the exclusion principle. I'm assuming that these six girls are fixed, and we're not considering any $6$ of the girls want this. That is, the number of ways to arrange the marriages such that $6$ girls don't marry a boy her age (call it $A$) is equal to the total number of marriage arrangements (call it $B$) minus the number of arrangements where those six girls do marry the boy her age (call it $C$). We know $B$ from the first part of the question and we want $A$. Now we say that those $6$ girls must marry the boy her age. How can that be represented by arranging them in line? Where must they stand in line if their marriage partner is already decided?