Maximum–Minimum Region Problems — notes, question types & practice · Set Theory & Venn Diagrams

01

Two bounds run everything

Region sizes cannot wander freely. Two rules bind them:

The union never exceeds the total: $n(A \cup B) \le N$ .
No region is negative.

Maximum questions push a region up; minimum questions push it down. Apply the two rules and the extremes fall out.

02

Maximum overlap of two sets

One circle can sit wholly inside the other.

\max n(A \cap B) = \min(n(A), n(B))

Worked: 120 like cricket and 100 like football in a group of 200. Maximum both $= 100$ — every football lover also likes cricket.

03

Minimum overlap of two sets

Stretch the union to its largest, $N$ . The overlap is then squeezed to its smallest.

\min n(A \cap B) = \max\left(0,\ n(A) + n(B) - N\right)

Worked: in a group of 100, 60 drink tea and 50 drink coffee. Minimum both $= 60 + 50 - 100 = 10$ .

Rule: If $n(A) + n(B) > N$ , at least $n(A) + n(B) - N$ people must sit in both. Counts cannot be negative, so keep the floor at 0.

04

Maximum 'neither' and minimum union

"Neither" is largest when the union is smallest, and the union cannot drop below its largest member set.

\max \text{neither} = N - \max(n(A), n(B))

Worked: 50 students, 30 like maths and 25 like physics. Maximum neither $= 50 - 30 = 20$ : everyone who likes physics also likes maths.

The same bound gives the minimum union of three sets: it is the largest of the three counts.

05

Three sets and the triple

The same two bounds fix the triple's limits.

Maximum triple $= \min(n(A), n(B), n(C))$ .
Minimum triple $= \max\left(0,\ n(A) + n(B) + n(C) - 2N\right)$ .

Why the minimum: a person outside the triple sits in at most two of the three sets. The singles total can therefore hide at most $2N$ memberships outside the middle, and whatever is left over must sit in all three.

Worked: everyone in a group of 150 likes at least one of tea (120), coffee (110) and milk (100). Minimum triple $= 330 - 300 = 30$ .

Watch: The formula needs the union fixed at $N$ . If people may like none, replace $N$ by the number who like at least one.

06

Extremes of the derived regions

Exactly one and at least two move opposite to the overlap.

Exactly one is smallest when "both" is largest: $n(A) + n(B) - 2\min(n(A), n(B))$ .
For three sets, at least two $= n(A) + n(B) + n(C) - n(A \cup B \cup C) - t$ , so it is smallest when the triple $t$ is largest.

Worked: 130 play cricket and 120 play football in a group of 200. Both can reach 120, so exactly one can drop to $250 - 240 = 10$ .

Note: After computing an extreme, check every region stays non-negative. If some region breaks, the extreme is not achievable.

07

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Maximum overlap (two sets or the triple)

How to spot it:

'Maximum possible number who like both / all three' with overlapping groups.

\min(n(A), n(B)) \quad \text{or} \quad \min(n(A), n(B), n(C))

Method

Take the smallest group size.
Nest the other sets inside it.
Check the union stays within the total.

Why it works:

No overlap can exceed the smallest set, and nesting achieves exactly that.

Try this

In a class of 90 students, 55 take Drawing and 40 take Music. What is the maximum number who take both?

Show solution

Maximum $= \min(55, 40)$ .
$= 40$ .

Answer

40

Type 2very common3 practice Q

Minimum overlap of two groups

How to spot it:

'At least how many like both' or 'minimum number who take both' with a fixed total.

\max(0,\ n(A) + n(B) - N)

Method

Add the two group sizes.
Subtract the total.
Keep a floor of 0 if the difference is negative.

Why it works:

Beyond the total, the two groups must share people, and the forced share is the minimum.

Try this

In a hostel of 90 students, 70 take tea and 35 take coffee, and every student takes at least one drink. What is the minimum number who take both?

Show solution

Minimum $= 70 + 35 - 90$ .
$= 15$ .

Answer

15

Type 3common3 practice Q

Maximum 'neither' or minimum union

How to spot it:

'Maximum number who play none' or 'minimum who play at least one' with a fixed class size.

\max \text{neither} = N - \max(n(A), n(B))

Method

Find the largest single group.
Shrink the union to that size by nesting.
Subtract the union from the total.

Why it works:

The union must contain its biggest member set, and nesting achieves that minimum.

Try this

In an office of 120 employees, 70 know typing, 60 know shorthand and 50 know data entry. What is the maximum number who know none of the three?

Show solution

Union at least $\max(70, 60, 50) = 70$ .
None at most $120 - 70 = 50$ .

Answer

50

Type 4common2 practice Q

Minimum triple overlap

How to spot it:

'At least how many like all three' when everyone is covered by three groups.

\max(0,\ n(A) + n(B) + n(C) - 2N)

Method

Add the three singles.
Subtract twice the total.
Keep a floor of 0.

Why it works:

A person outside the triple holds at most two memberships, so extras beyond $2N$ pile into the middle.

Try this

In a group of 200 people, every person likes at least one of tea, coffee and milk. 150 like tea, 140 like coffee and 130 like milk. What is the minimum number who like all three?

Show solution

Singles: $150 + 140 + 130 = 420$ .
$420 - 2 \times 200 = 20$ .

Answer

20

Type 5occasional2 practice Q

Extremes of exactly one and at least two

How to spot it:

'Minimum who play exactly one game' or 'minimum who play at least two games'.

\text{exactly one} = n(A) + n(B) - 2\,n(A \cap B)

Method

Push the overlap to its extreme (largest or smallest).
Compute the region from the overlap.
Check all regions stay non-negative.

Why it works:

These regions fall as the overlap rises, so their extremes sit at the overlap's extremes.

Try this

In a group of 150 people, 90 like tea and 80 like coffee. What is the minimum number who like exactly one drink?

Show solution

Both at most $\min(90, 80) = 80$ .
Exactly one: $170 - 160 = 10$ .

Answer

10

08

Formula sheet

Maximum overlap

\max n(A \cap B) = \min(n(A), n(B))

One set wholly inside the other.

Minimum overlap

\min n(A \cap B) = \max(0,\ n(A) + n(B) - N)

$N$ is the size of the universal set.

Maximum neither

N - \max(n(A), n(B))

Shrink the union to its largest member set.

Minimum triple

\max(0,\ n(A) + n(B) + n(C) - 2N)

Union fixed at $N$; everyone covered.

Maximum triple

\min(n(A), n(B), n(C))

All three sets nested.

09

Shortcuts that save time

⚡ Push the smaller set inside the larger

For a maximum overlap, nest the circles. The smaller group sits entirely inside the bigger one.

Example

In a class of 45 students, 28 play cricket and 22 play football. What is the maximum number who play both?

Show solution

Maximum $= \min(28, 22)$ .
$= 22$ .

Answer

22

⚡ Min overlap: stretch the union to its limit

Let the two circles cover as many people as the total allows. The forced sharing is the answer.

Example

In a group of 80 people, 50 like tea and 45 like coffee. What is the minimum number who like both?

Show solution

$50 + 45 - 80$ .
$= 15$ .

Answer

15

⚡ Leftover demand forces the triple

Singles beyond $2N$ cannot fit outside the middle. Subtract $2N$ from the singles sum.

Example

In a group of 120 people where everyone likes at least one drink, 110 like tea, 90 like coffee and 80 like milk. What is the minimum number who like all three?

Show solution

Singles: $280$ .
$280 - 2 \times 120 = 40$ .

Answer

40

10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Calling $n(A) + n(B) - N$ the maximum overlap — that is the minimum when the total is fixed.

Mistake 02

Allowing a negative minimum — write $\max(0,\ n(A) + n(B) - N)$ ; counts cannot go below zero.

Mistake 03

Comparing with the wrong total — the bound uses the universe size, not the sum of the groups.

Mistake 04

Assuming the minimum overlap is always 0 — with a tight total it is forced above zero.

Mistake 05

Maximising 'neither' by shrinking both circles — only the largest set must stay whole.

Mistake 06

Skipping the achievability check — every Venn region must stay non-negative at your extreme.

11

Quick revision

Read this the night before the exam.

Max both $= \min(n(A), n(B))$ ; max triple $= \min$ of the three.
Min both $= \max(0,\ n(A) + n(B) - N)$ .
Max neither $= N - \max(n(A), n(B))$ .
Min triple $= \max(0,\ n(A) + n(B) + n(C) - 2N)$ .
Exactly one and at least two are extreme at the overlap's extremes.
Always re-check that no region goes negative.

12

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.