Higher Mathematics · Classes 9-10 · Chapter 1
Number of Elements of Finite Sets and Sets in Solving Real-Life Problems.
Thirty five students speak English, forty speak Bangla, and there are only fifty in the class. Where did a total of seventy five come from? Everyone who speaks both was counted twice. That one thought drives the chapter's most examined skill: putting numbers into a Venn diagram.
1. Counting disjoint sets
The number of elements of a finite set A is written n(A). Proposition 8: if A and B are disjoint finite sets, n(A ∪ B) = n(A) + n(B). With n(A) = p and n(B) = q, number the elements of A from 1 to p and those of B from p + 1 to p + q; so A ∪ B ∼ Jp+q.
The same holds for more sets. If they are pairwise disjoint, n(A ∪ B ∪ C) = n(A) + n(B) + n(C), and with four, n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D). Nobody overlaps anybody, so nobody is counted twice.
2. Any two sets
Proposition 9: for any finite sets A and B, n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
A has 7 dots and B has 5. Change how many are in both, and watch how far the plain sum drifts from the real count.
Where does the sum go wrong?
7 and 5 are not the book's numbers; they are there for you to play with.
The proof uses the three pieces of the diagram. A \ B, A ∩ B and B \ A are disjoint, so Proposition 8 applies: n(A) = n(A \ B) + n(A ∩ B), n(B) = n(B \ A) + n(A ∩ B), and n(A ∪ B) = n(A \ B) + n(A ∩ B) + n(B \ A). Put n(A \ B) and n(B \ A) from the first two into the third and the formula falls out.
On the way you get a working formula too: n(A \ B) = n(A) − n(A ∩ B). That is the number in the “only A” region.
For a complement, subtract from the universal set: n(A′) = n(U) − n(A). In Example 16, 15 of the 35 girls like running, so n(J′) = 35 − 15 = 20 do not.
3. From words to symbols
The wording decides which number is wanted. “At least one” is n(A ∪ B). “Both” is n(A ∩ B). “Only one” is n(A ∪ B) − n(A ∩ B). “Neither” is n(U) − n(A ∪ B). And “everyone likes at least one” means A ∪ B = U, so the neither box holds zero.
Match the meaning
4. Two-set problems
Example 15: of 50 people, 35 speak English, 25 speak both English and Bangla, and everyone speaks at least one. So n(E ∪ B) = 50, and 50 = 35 + n(B) − 25, which gives 40 Bangla speakers. Only Bangla: 40 − 25 = 15.
The book's activity has three of the same shape. Of 30, 20 like football and 15 chess, and everyone likes at least one: both = 20 + 15 − 30 = 5. 50 speak Bangla, 20 English, 10 both: at least one = 50 + 20 − 10 = 60. Of 50, 29 take Biology, 24 Higher Mathematics, 11 both: neither = 50 − (29 + 24 − 11) = 8.
Know four numbers and the fifth follows, because n(U) = n(A) + n(B) − n(A ∩ B) + neither. Choose which one is unknown and enter the rest. The overlap is filled first, then the other three boxes.
Fill the overlap first
Nothing you type goes anywhere. The page is yours. The diagram is drawn by hand, not to scale.
Example 17: of 24 students, 18 play basketball (B) and 12 volleyball (V). Nobody says how many play both. Slide x and see which values push a box below zero.
How small, how large can x be?
The circles are drawn by hand, not to scale. They slide together as x grows only to show the overlap widening.
Now write down which values of x survived.
In Example 17 the overlap is unknown, and the question is how small and how large it can be. Of 24 students, 18 like basketball and 12 volleyball. With n(B ∩ V) = x, n(B ∪ V) = (18 − x) + x + (12 − x) = 30 − x. The union cannot be bigger than U, so x is smallest when B ∪ V = U: 30 − x = 24, x = 6. And x is largest when V ⊂ B: x = n(V) = 12.
5. Three sets
For three finite sets the formula is one step longer: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). The triple overlap is added three times with the single sets, taken away three times with the pairs, so it goes back in once at the end.
Fill a three-set diagram from the inside out. First the triple overlap. Then each “exactly two” region: its pairwise overlap minus the triple. Then each “only one” region: its set minus everything else inside it. Last, neither = n(U) − n(union).
The language institute activity has 100 students: French 42, German 30, Spanish 28; French and Spanish 10, German and Spanish 8, German and French 5, all three 3. Union = 42 + 30 + 28 − 10 − 8 − 5 + 3 = 80, so 20 take none. “Exactly two” = (sum of the pairwise overlaps) − 3 × (triple) = 23 − 9 = 14. “Exactly one” = union − exactly two − all three = 80 − 14 − 3 = 63. Fill it step by step.
Start at the centre
Exercise 17: 30 students; Economics 19, Geography 17, Civics 11; Economics and Geography 12, Civics and Geography 4, Economics and Civics 7, all three 3. Union = 19 + 17 + 11 − 12 − 4 − 7 + 3 = 27, so 3 took none.
Exercise 27 runs the other way: the triple is unknown and the number taking none is given. 100 students; football 42, cricket 46, chess 39; football and cricket 13, cricket and chess 14, football and chess 12; 7 play none. The union is 93 = 127 − 39 + x, so x = 5 play all three. Exactly two = 39 − 15 = 24, at least two = 24 + 5 = 29, exactly one = 93 − 29 = 64.
Percentages work the same way (Exercise 25): Bichitra 60%, Sandhani 50%, Purbani 50%; pairwise 30%, 30%, 20%; all three 10%. At least one = 60 + 50 + 50 − 30 − 30 − 20 + 10 = 90%, none 10%, exactly two = 80 − 30 = 50%.
Enter the counts and the calculator shows the adding, the subtracting and the triple added back. If the triple is the unknown, change “find”.
Add, subtract, then add the triple back
Nothing you type goes anywhere. The page is yours.
6. Unknowns inside the diagram
Example 16: each of 35 girls likes at least one of running (J), swimming (S) and dancing (D). The diagram shows only running 2, running and swimming but not dancing 7, all three 4, only swimming 2, swimming and dancing but not running x, and only dancing 2x. Everyone outside running makes n(J′) = 20, so 2x + x + 2 = 20 and x = 6. If y like running and dancing but not swimming, the 15 runners give y + 4 + 7 + 2 = 15, so y = 2.
Exercise 6: only A = 3x, A ∩ B = x, only B = 2x + 8, and n(A) = n(B). Then 4x = 3x + 8 and x = 8. n(A ∪ B) = 3x + x + 2x + 8 = 56, and n(B \ A) = 2x + 8 = 24.
Exercise 18: three circles in a row, with A and C not meeting. Only A 8, A ∩ B = x, only B 6, B ∩ C = 4, only C = y. n(A ∩ B) = n(B ∩ C) gives x = 4; n(B ∩ C′) = n(A′ ∩ C) gives x + 6 = 4 + y, so y = 6. The boxes are disjoint, so n(U) = 8 + 4 + 6 + 4 + 6 = 28.
Exercise 19: n(U) = 50; only A 2x, the A ∩ B-only box x + 1, only B x − 1, the A ∩ C-only box 2, all three 3, the B ∩ C-only box 0, only C x + 5. Adding the seven boxes: 5x + 10 = 50, so x = 8. Then n(B ∩ C′) = 9 + 7 = 16, n(A′ ∩ B) = 7 + 0 = 7, and n(A ∩ B ∩ C′) = 9.
Number sets are counted the same way (Exercise 7): U = the positive integers, A = {x : x ≥ 5}, B = {x : 5x < 12} = {1, 2}. A ∩ B = ∅, so n(A ∩ B) = 0. A′ = {1, 2, 3, 4}, so A′ ∪ B = {1, 2, 3, 4}, with 4 elements.
7. Check yourself
The first comes from Proposition 8.
Question 1
If n(A) = p, n(B) = q and A ∩ B = ∅, what is n(A ∪ B)?
In the second, notice the words “at least one”.
Question 2
Among some people, 50 speak Bangla, 20 speak English and 10 speak both. How many speak at least one of the two languages?
The last is the question of Example 17.
Question 3
Of 24 students, 18 play basketball and 12 play volleyball. What is the smallest possible number who play both?
The whole note on eight cards.
Eight cards
Source: Higher Mathematics, Classes 9-10 (NCTB), Chapter 1 “Set and Function”, the “Number of Elements of Finite Sets and Sets in Solving Real-Life Problems” part.