Largest and smallest possible counts

Secondary 4

When only the totals are known

Sometimes only $n(\xi)$, $n(A)$ and $n(B)$ are known. Then the overlap $n(A\cap B)$ is not fixed: it can take a range of whole-number values, and every other region count depends on it. Because $n(A\cup B)=n(A)+n(B)-n(A\cap B)$, a larger overlap means a smaller union.

Call the set with fewer elements the smaller set. The extremes come from three pictures.

Smaller set $B$ inside $A$: largest overlap.

Disjoint: smallest overlap when $n(A)+n(B)\le n(\xi)$.

No outside region: smallest overlap when $n(A)+n(B)>n(\xi)$.

Picture	$n(A\cap B)$	$n(A\cup B)$	$n((A\cup B)')$
Smaller set inside the larger set	largest: $n(\text{smaller set})$	smallest: $n(\text{larger set})$	largest: $n(\xi)-n(\text{larger set})$
Disjoint, if $n(A)+n(B)\le n(\xi)$	smallest: $0$	largest: $n(A)+n(B)$	smallest: $n(\xi)-n(A)-n(B)$
$A\cup B=\xi$, if $n(A)+n(B)>n(\xi)$	smallest: $n(A)+n(B)-n(\xi)$	largest: $n(\xi)$	smallest: $0$

Why: the overlap is part of both sets, so it can never contain more elements than the smaller set, and it reaches that size only when the smaller set lies inside the larger one. The union can never contain more than $n(\xi)$ elements; if the two sets together have more elements than $\xi$, some elements must be shared.

Method

Identify the smaller set.
For the largest overlap, draw the smaller set inside the larger set and fill in every region.
For the smallest overlap, compare $n(A)+n(B)$ with $n(\xi)$: draw the sets apart if the sum is at most $n(\xi)$; otherwise leave no outside region and the overlap is $n(A)+n(B)-n(\xi)$.
Read the asked count from the diagram that makes it largest or smallest.

Example

$n(\xi)=30$, $n(A)=18$ and $n(B)=16$. Find the largest and smallest possible values of $n(A\cap B)$ and of $n((A\cup B)')$.

The smaller set is $B$. Largest overlap: $B\subseteq A$, so $n(A\cap B)=16$, $n(A\cap B')=18-16=2$ and $n((A\cup B)')=30-18=12$.

Smallest overlap: $18+16=34>30$, so the sets cannot be disjoint. With no outside region, $n(A\cap B)=34-30=4$, giving $14$, $4$ and $12$ in the three regions and $0$ outside.

Largest $n(A\cap B)=16$

Smallest $n(A\cap B)=4$

So $4\le n(A\cap B)\le 16$, and $n((A\cup B)')$ is largest at $12$ and smallest at $0$.

Check

Every region count must be a whole number that is not negative, and the four counts must add up to $n(\xi)$: $2+16+0+12=30$ and $14+4+12+0=30$. A common error is to give the larger set as the largest overlap, or to draw the sets apart when $n(A)+n(B)>n(\xi)$, which makes the outside count negative.

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Picture	\(n(A\cap B)\)	\(n(A\cup B)\)	\(n((A\cup B)')\)
Smaller set inside the larger set	largest: \(n(\text{smaller set})\)	smallest: \(n(\text{larger set})\)	largest: \(n(\xi)-n(\text{larger set})\)
Disjoint, if \(n(A)+n(B)\le n(\xi)\)	smallest: \(0\)	largest: \(n(A)+n(B)\)	smallest: \(n(\xi)-n(A)-n(B)\)
\(A\cup B=\xi\), if \(n(A)+n(B)>n(\xi)\)	smallest: \(n(A)+n(B)-n(\xi)\)	largest: \(n(\xi)\)	smallest: \(0\)