In the previous section, we saw some interesting relations between three sets. In this section, we will see some practical problems involving three sets.
In a previous section 1.9, we saw the practical problems involving two sets. We saw that some students wanted to be in both cricket team and football team.
We derived Eq.1.1: n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Using the equation, we solved the situation effectively.
• Now suppose that, a hockey coach also wants to make a team.
♦ Some students will want to be in both cricket team and football team.
♦ Some students will want to be in both cricket team and hockey team.
♦ Some students will want to be in both football team and hockey team.
♦ Some students will want to be in all three teams.
• Here also, we can derive an equation similar to the Eq.1.1
♦ It can be derived in 7 steps:
1. We want to find n(A ∪ B ∪ C)
• Let us treat B ∪ C as one unit. Then we can write:
n(A ∪ B ∪ C) = n[A ∪ (B ∪ C)]
♦ Recall that A ∪ (B ∪ C) = (A ∪ B) ∪ C = A ∪ B ∪ C
♦ See figs.1.7 and 1.8 in section 1.5 for proof
2. Now (B ∪ C) occupies a position similar to 'B' in Eq.1.1
• So we can write:
n[A ∪ (B ∪ C)] = n(A) + n(B ∪ C) - n[A ∩ (B ∪ C)]
3. The right side has three terms. We apply Eq.1.1 to the second term. We get:
n[A ∪ (B ∪ C)] = n(A) + n(B) + n(C) - n(B ∩ C) - n[A ∩ (B ∪ C)]
4. Now the right side has 5 terms. We apply the 'distributive law of intersection' to the fifth term.
• Then the fifth term becomes:
n[(A ∩ B) ∪ (A ∩ C)]
5. We apply Eq.1.1 to this modified fifth term. We get:
n[(A ∩ B) ∪ (A ∩ C)] = n(A ∩ B) + n(A ∩ C) - n[(A ∩ B) ∩ (A ∩ C)]
6. The last term on the right side is simply: n(A ∩ B ∩ C)
• So we can write:
n[(A ∩ B) ∪ (A ∩ C)] = n(A ∩ B) + n(A ∩ C) - n(A ∩ B ∩ C)
7. So the fifth term in (3) can be replaced. The modified equation is:
n[A ∪ (B ∪ C)] = n(A) + n(B) + n(C) - n(B ∩ C) - n(A ∩ B) - n(A ∩ C) + n(A ∩ B ∩ C)
• Rearranging this, we get Eq.1.2:
n[A ∪ B ∪ C] = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)
• There is a definite pattern in this equation:
♦ The individual numbers n(A), n(B) and n(C) are added.
♦ The intersecting pairs n(A ∩ B), n(B ∩ C) and n(A ∩ C) are subtracted.
♦ The overall intersection n(A ∩ B ∩ C) is added.
Let us see a solved example
Solved example 1.69
A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports, how many received medals in exactly two of the three sports ?
Solution:
1. Given that: n(F) = 38, n(B) = 15 and n(C) = 20
• Also given that: n(F ∪ B ∪ C) = 58 and n(F ∩ B ∩ C) = 3
2. Applying Eq.1.2, we get:
n(F ∪ B ∪ C) = n(F) + n(B) + n(C) - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + n(F ∩ B ∩ C)
3. Substituting the known values, we get:
58 = 38 + 15 + 20 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C) + 3
⇒ 58 = 76 - n(F ∩ B) - n(B ∩ C) - n(F ∩ C)
⇒ n(F ∩ B) + n(B ∩ C) + n(F ∩ C) = 18
4. The intersection pairs are added three times in the above result in (3)
• That sum in the Venn diagram below is: (a+d) + (b+d) + (c+d)
 |
| Fig.1.30 |
5. We want (a+b+c)
• It can be obtained as:
a+b+c = (a+d) + (b+d) + (c+d) -3d
6. Thus we get:
a+b+c = n(F ∩ B) + n(B ∩ C) + n(F ∩ C) - 3d
⇒ a+b+c = 18 - (3 × 3) = 9
Solved example 1.70
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read
newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,
8 read both T and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution:
1. Given that: n(H) = 25, n(T) = 26 and n(I) = 26
• Also given that: n(H ∩ I) = 9, n(H ∩ T) = 11, n(T ∩ I) = 8 and n(H ∩ T ∩ I) = 3
2. This problem can be easily solved using Venn diagrams. The various portions in the Venn Venn diagram can be filled up in 7 steps:
I. Given n(H ∩ T ∩ I) = 3
• So 3 comes in the central portion. This is marked as I in fig.1.31(a) below:
 |
| Fig.1.31 |
II. Given n(H ∩ I) = 9
• So the overlap between H and I is 9
• But 3 is already present in this overlap.
♦ So the remaining portion of the overlap is (9 - 3) = 6
• This is marked as II in fig.1.31(a) above.
III. Given n(H ∩ T) = 11
• So the overlap between H and T is 11
• But 3 is already present in this overlap.
♦ So the remaining portion of the overlap is (11 - 3) = 8
• This is marked as III in fig.1.31(a) above.
IV. Given n(T ∩ I) = 8
• So the overlap between T and I is 8
• But 3 is already present in this overlap.
♦ So the remaining portion of the overlap is (8 - 3) = 5
• This is marked as IV in fig.1.31(a) above.
V. Given n(H) = 25
• So the circle H will enclose 25
• But 8, 3 and 6 are already present inside H.
♦ So the extra needed is: [25 - (8+3+6)] = 8
• This is marked as V in fig.1.31(b) above.
VI. Given n(T) = 26
• So the circle T will enclose 26
• But 8, 3 and 5 are already present inside T.
♦ So the extra needed is: [26 - (8+3+5)] = 10
• This is marked as VI in fig.1.31(b) above.
VII. Given n(I) = 26
• So the circle I will enclose 26
• But 6, 3 and 5 are already present inside I.
♦ So the extra needed is: [26 - (6+3+5)] = 12
• This is marked as VII in fig.1.31(b) above.
3. Now the Venn diagram is completely filled up. We can answer the questions.
(i) the number of people who read at least one of the newspapers.
• The answer will be n(H ∪ T ∪ I)
♦ To find this number, we have to add all items inside the three circles.
♦ So we get: n(H ∪ T ∪ I) = (8+8+6+3+10+5+12) = 52
(ii) the number of people who read exactly one newspaper.
• For this, we need to add the non-intersecting portions of the three circles
♦ We get: (8+12+10) = 30
Solved example 1.71
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution:
1. Given that: n(A) = 21, n(B) = 26 and n(C) = 29
• Also given that: n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14 and n(A ∩ B ∩ C) = 8
2.
This problem can be easily solved using Venn diagrams. The various
portions in the Venn Venn diagram can be filled up in 7 steps:
I. Given n(A ∩ B ∩ C) = 8
• So 8 comes in the central portion. This is marked as I in fig.1.32(a) below:
 |
| Fig.1.32 |
• So the overlap between A and B is 14
• But 8 is already present in this overlap.
♦ So the remaining portion of the overlap is (14 - 8) = 6
• This is marked as II in fig.1.32(a) above.
III. Given n(C ∩ A) = 12
• So the overlap between C and A is 12
• But 8 is already present in this overlap.
♦ So the remaining portion of the overlap is (12 - 8) = 4
• This is marked as III in fig.1.32(a) above.
IV. Given n(B ∩ C) = 14
• So the overlap between B and C is 14
• But 8 is already present in this overlap.
♦ So the remaining portion of the overlap is (14 - 8) = 6
• This is marked as IV in fig.1.32(a) above.
V. Given n(A) = 21
• So the circle A will enclose 21
• But 8, 4 and 6 are already present inside A.
♦ So the extra needed is: [21 - (8+4+6)] = 3
• This is marked as V in fig.1.32(b) above.
VI. Given n(B) = 26
• So the circle B will enclose 26
• But 6, 8 and 6 are already present inside B.
♦ So the extra needed is: [26 - (6+8+6)] = 6
• This is marked as VI in fig.1.31(b) above.
VII. Given n(C) = 29
• So the circle C will enclose 29
• But 4, 8 and 6 are already present inside C.
♦ So the extra needed is: [29 - (4+8+6)] = 11
• This is marked as VII in fig.1.32(b) above.
3. Now the Venn diagram is completely filled up. We can answer the question.
How many like product C only?
• The answer will be the non-intersecting portion of C. It is equal to 11
In
the next
chapter, we will see the Relations and Functions.
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