How to find the size of the complement of a three-set intersection

Expert intro: This is a neat counting question. The fastest way is to list the sets inside the universal set, find the overlap, then take the complement.

Detailed walkthrough

Using the same universal set

$U=\{41,42,43,44,45,46,47,48,49\},$

we have:

• $P=\{49\}$, the square numbers
• $Q=\{41,43,45,47,49\}$, the odd numbers
• $R=\{42,45,48\}$, the multiples of 3

Now look at the intersection:

$P\cap Q\cap R.$

• $49$ is in $P$ and $Q$, but not in $R$.
• None of the multiples of 3 in $U$ are square numbers.

So

$P\cap Q\cap R=\varnothing.$

Therefore its complement is the whole universal set:

$(P\cap Q\cap R)'=U.$

Hence

$\boxed{n\big((P\cap Q\cap R)'\big)=9}.$

💡 Pitfall guide

Do not confuse $P\cap Q\cap R$ with $P\cap(Q\cap R)$ as if one set can be ignored. Intersections must satisfy all three conditions at once. Also, once the triple intersection is empty, its complement is every element in $U$, not just the elements from one or two of the sets.

🔄 Real-world variant

If the universal set were changed to include $36$, then $36$ would belong to $P$ and $R$, but not to $Q$ because it is even. So the triple intersection would still be empty. To get a non-empty triple intersection, you need an element that is simultaneously square, odd, and divisible by 3, which is much rarer.

🔍 Related terms

complement, cardinality, intersection