For any two sets A and B:

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

This formula accounts for elements that appear in both sets (the intersection) so they are not counted twice.

In a seminar of 40 people, 16 use gas for cooking, 25 use charcoal, and 6 use neither. How many people use both gas and charcoal?

Solution

Let:

• μ = {all people at the seminar}, so n(μ) = 40
• G = {people who use gas}, n(G) = 16
• C = {people who use charcoal}, n(C) = 25
• n(G ∪ C)' = 6 (use neither)

First, find how many use at least one:

$n(G \cup C) = n(\mu) - n(G \cup C)' = 40 - 6 = 34$

Now apply the formula:

$n(G \cup C) = n(G) + n(C) - n(G \cap C)$

$34 = 16 + 25 - n(G \cap C)$

$n(G \cap C) = 41 - 34 = 7$

Answer: 7 people use both gas and charcoal.

A community has 500 people. There are 300 in a climate action group and 250 in a poverty alleviation group. If 180 people are in both groups, find:

(a) Number in climate action only (b) Number in poverty alleviation only (c) Number in at least one group

Solution

Given: n(μ) = 500, n(C) = 300, n(P) = 250, n(C ∩ P) = 180

(a) Climate action only = n(C ∩ P'):

$n(C) = n(C \cap P) + n(C \cap P')$

$300 = 180 + n(C \cap P')$

$n(C \cap P') = 300 - 180 = 120$

(b) Poverty alleviation only = n(C' ∩ P):

$n(P) = n(C \cap P) + n(C' \cap P)$

$250 = 180 + n(C' \cap P)$

$n(C' \cap P) = 250 - 180 = 70$

(c) At least one group = n(C ∪ P):

$n(C \cup P) = n(C) + n(P) - n(C \cap P)$

$n(C \cup P) = 300 + 250 - 180 = 370$

Answers: (a) 120 people, (b) 70 people, (c) 370 people

The universal set μ contains all elements under discussion. The complement A' contains elements in μ that are not in A.

Relationship:

$n(\mu) = n(A) + n(A')$

• Always identify what each set represents
• Draw a Venn diagram when helpful to visualize the problem
• Use the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B) for two-set problems
• Check that your answer makes sense (e.g., cannot have negative numbers)

In Tanzania, market vendors often use this skill when tracking inventory. For example, a shopkeeper may know that 45 customers bought rice and 30 bought beans last month, with 15 buying both. Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B), they can calculate that 60 different customers bought rice or beans (or both), helping them plan restocking and understand which combinations of items are popular.