Revision notes
Use Venn diagrams to solve counting problems with two or three sets — start in the middle and work outwards.
In Emeka's class of 40, 25 like jollof rice, 18 like fried rice and 10 like both. If you just add 25 + 18 you get 43 — more than the whole class! A Venn diagram shows exactly where the double counting happens.
A rectangle stands for the universal set ξ, and a circle for each set. Where the circles overlap is A ∩ B — the elements in both.
Always fill the overlap first. In the hook, 10 like both. Then jollof only = 25 − 10 = 15 and fried rice only = 18 − 10 = 8.
Students in at least one circle: 15 + 10 + 8 = 33. So 40 − 33 = 7 like neither — they go outside the circles but inside the rectangle.
🔑 Key idea: Start from the middle (the overlap) and work outwards.
Adding n(A) + n(B) counts the overlap twice, so we subtract it once: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
If the overlap is unknown, call it x. With 40 students, 25 + 18 − x + 7 = 40 gives x = 10.
🔑 Key idea: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
With three circles there is one centre region (all three) and three 'two sets only' regions.
Fill the centre first, then each 'two only' region (subtract the centre), then each 'one only' region. For example, if 12 offer Physics and Chemistry and 4 of them also offer Biology, then Physics and Chemistry only = 12 − 4 = 8.
🔑 Key idea: Centre → two-only regions → one-only regions → outside.
In a class of 48 students, 22 offer Igbo, 29 offer Yoruba and 9 offer neither. How many offer both?
Step 1: Students in at least one group
48 − 9 = 39
Step 2: Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
39 = 22 + 29 − x
x = 51 − 39 = 12
Why? The 51 counts the 'both' students twice; the extra over 39 is exactly the overlap.
Answer: 12
Ready to check you've got it? Scroll up and work through the lesson: the interactive visuals, practice questions and mastery check are all there.