Union and Intersection Formulae and Concepts
In this article, we will study about the concepts of Union and Intersection of two or more sets in much more detail.
Union and Intersection formulae for two and three sets
Formula 1
If A and B are finite sets then:
n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
Q. In a college there are 30 teachers. 22 teachers teach English and 6 teachers teach both Maths and English. How many teachers must be teaching Maths?
Explanations :
Let E be the set of teachers who teach English and M be the set of teachers who teach Maths.
It’s given that: n (E ∪ M) = 30, n (E) = 22 and n (E ∩ M) = 6
We know that, n (E ∪ M) = n (E) + n (M) – n (E ∩ M)
or 30 = 22 + n (M) – 6
or n (M) = 14
Hence 14 teachers teach Maths.
Let E be the set of teachers who teach English and M be the set of teachers who teach Maths.
We can see from the above Venn diagram, that teachers who teach only English = 22 - 6 = 16
So, teachers who teach Maths = Total number of teachers - Teachers who teach only English = 30 - 16 = 14
Formula 2
If A, B and C are finite sets then:
n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C)
Q. In a school 38 students passed in mathematics, 15 students passed in Physics and 20 students passed in Chemistry. If only three students passed in all the three subjects and there are total 58 students in the school then how many students passed in exactly two of the three subjects?
Explanations :
Let A, B and C be the set of students passed in Mathematics, Physics and Chemistry respectively.
It’s given that: n (A) = 38, n (B) = 15, n (C) = 20, n (A ∪ B ∪ C) = 58 and n (A ∩ B ∩ C) = 3
We know that, n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (A ∩ C) – n (B ∩ C) + n (A ∩ B ∩ C)
or 58 = 38 + 15 + 20 – {n (A ∩ B) + n (A ∩ C) + n (B ∩ C)} + 3
or n (A ∩ B) + n (A ∩ C) + n (B ∩ C) = 18
- Here, ‘a’ denotes the number of students who passed in Mathematics and Physics only,
- ‘b’ denotes the number of students who passed in Mathematics and Chemistry only,
- ‘c’ denotes the number of students who passed in Physics and Chemistry only and
- 3 students passed in all the three subjects.
Now, n (A ∩ B) + n (A ∩ C) + n (B ∩ C) = 18
or (a + 3) + (b + 3) + (c + 3) = 18
or a + b + c = 9
Hence, 9 students passed in exactly two of the three subjects.
Using the Venn diagram method:
58 = (35 - a - b) + (12 - a - c) + (17 - b - c) + a + b + c + 3
or 58 = 35 + 12 + 17 + 3 - a - b - c
or a + b + c = 9
Hence, 9 students passed in exactly two of the three subjects.
Keywords related to Set Theory
While solving Set Theory questions, you will often come across some keywords that you must understand, or else you won’t be able to understand the problems correctly and make unforced errors.
We will study about: and, or, only, at least, at most, no/not
Have a look at the following Venn diagram and try to understand what these keywords mean in Set Theory:
Only
A = P + X + Y + T
A only = P
B = Q + X + Z + T
B only = Q
C = R + Y + Z + T
C only = R
And
And implies Intersection, i.e. ∩.
A AND B = X + T
A AND B only = X
B AND C = Z + T
B AND C only = Z
A AND C = Y + T
A AND C only = Y
Or
Or implies Union, i.e. ∪.
A OR B = P + Y + X + T + Q + Z
A OR B only = P + Q + X
A only OR B only = P + Q
B OR C = X + Q + T + Z + Y + R
B OR C only = Q + R + Z
B only OR C only = Q + R
A OR C = P + X + Y + T + R + Z
A OR C only = P + R + Y
A only OR C only = P + R
Not/None
None of Three = (A + B + C)′, i.e. (A ∪ B ∪ C)′
Not A OR B = (A OR B)′ = R + (A + B + C)′
Not B OR C = (B OR C)′ = P + (A + B + C)′
Not A OR C = (A OR C)′ = Q + (A + B + C)′
At least and At most
At least two means 2 or more, i.e. not less than 2. As we have only three sets A, B and C, so here ‘at least’ means 2 or 3.
At least Two = X + Y + Z + T
At most two means 2 or less (2, 1, or 0), i.e. not more than 2.
At most Two = Total – T = U - T = P + Q + R + X + Y + Z + (A ∪ B ∪ C)′
This will become much more clear with a few examples.
Q. The following Venn diagram represent the number of people can speak three languages viz. Tamil, Telugu or English.
Find number of people, who can speak:(i) Tamil AND Telugu
(ii) Tamil AND Telugu only
(iii) Tamil OR Telugu only
(iv) Tamil only OR Telugu only
(v) Not Tamil OR Telugu
(vi) Either Tamil OR Telugu
(vii) At least two languages
(viii) At most two languages
Explanation:
(i) Number of people who can speak Tamil AND Telugu = 12+9 = 21
(ii) Number of people who can speak Tamil and Telugu only = 12
(iii) Number of people who can speak Tamil OR Telugu only = 35 + 12 + 24 = 71
(iv) Number of people who can speak Tamil only OR Telugu only = 24 + 35 = 59
(v) Number of people who can speak Not Tamil OR Telugu = (Tamil OR Telugu)′ = 27
(vi) Number of people who can speak either Tamil or Telugu = 35 + 12 + 24 + 9 + 13 + 19 = 112
(vii) Number of people who can speak at least 2 languages = 12 + 13 + 19 + 9 = 53
(viii) Number of people who can speak at most 2 languages = Total – 9 or 35 + 24 + 27 + 12 + 19 + 13 = 130
Q. In a class of 150 students, 50 students passed in mathematics, 40 students failed only in Hindi and 20 students failed in both the subjects. Find:
(i) How many students passed in both the subjects?
(ii) How many students passed exactly in one subject?
(iii) How many students failed in atleast one of the subjects?
Explanation:
Total number of students = 150
Number of students passed in Mathmatics = 50
Number of students failed in Mathmatics = 150 – 50 = 100
Number of students passed in Hindi = Total number of students – number of students failed in Hindi = 150 – 60 = 90
(i) The Venn diagram based on the given information is shown below:
Number of students who passed in both the subjects = Number of students who did not fail = 150 – (80 + 20 + 40) = 10
(ii) The Venn diagram based on the given information is shown below:
Required answer = Number of students who passed in Mathematics only (but did fail in Hindi) + number of students who passed in Hindi only (but did fail in Mathematics) = 40 + 80 = 120
(iii) The number of people who failed in at least one of the subjects = number of students who failed in Maths (but passed in Hindi) + the number of students who failed in Hindi (but passed in Maths) + the number of students who failed in both subjects = 80 + 40 + 20 = 140.
Or this can be obtained by subtracting the students who passed in both the subjects from the total number of students = 150 – 10 = 140
Q. The diagram gives the number of candidates who failed in Chemistry, Electronics and English tests in college. The total number of candidates that appeared in these 3 tests was 600.
(i) What is the percentage of candidates who failed in at least two subjects?
(ii) What is the total number of passed candidates?
(iii) What is the number of candidates who passed in at least one subject?
(iv) What is the number of candidates passed in only Chemistry?
(v) What is the number of candidates passed in only English?
(vi) What is the number of candidates passed in only Electronics?
(vii) What is the number of candidates passed in both Chemistry and Electronics?
(viii) What is the number of candidates passed in both Electronics and English?
(ix) What is the number of candidates passed in both Chemistry and English?
(x) What is the number of candidates passed in at least two subjects?
(xi) What is the number of candidates passed in at most two subjects?
(xii) What is the number of candidates passed in at most one subject?
Explanation:
(i) Candidates failed in at least two subjects = 15 + 10 + 22 + 28 = 75
Required percentage = $\frac{75}{600}$ × 100 = 12.5%
(ii) Total failed candidates = 27 + 65 + 90 + 15 + 10 + 22 + 28 = 257
Total passed candidates passed in all the three subjects = 600 - 257 = 343
(iii) Number of candidates who passed in at least one subject = Total number of candidates appeared in test – Total number of candidates who failed in all the three subjects = 600 – 10 = 590
(iv) Number of candidates passed in only Chemistry = candidates failed in rest two subjects (i.e. Electronics and English) = 22
(v) Number of candidates passed in only English = candidates failed in rest two subjects (i.e. Electronics and Chemistry) = 15
(vi) Number of candidates passed in only Electronics = candidates failed in rest two subjects (i.e. English and Chemistry) = 28
(vii) Number of candidates failed in both Chemistry and Electronics = 15 + 10 + 27 + 65 + 22 + 28 = 167
Number of candidates passed in both Chemistry and Electronics = 600 - 167 = 437
(viii) Number of candidates failed in both Electronics and English = 15 + 10 + 90 + 65 +22 + 28 = 240
Number of candidates passed in both Electronics and English = 600 - 240 = 360
(ix) Number of candidates failed in both Chemistry and English = 27 + 28 + 90 + 10 + 22 + 15 = 192
Number of candidates passed in both Chemistry and English = 600 - 192 = 408
(x) Number of candidates passed in at least two subjects = candidates passed in at least two subjects + candidates passed in at least three subjects = 27 + 65 + 90 + 343 = 525
(xi) Number of candidates passed in at most two subjects = candidates passed in two subjects + candidates passed in one subject + candidates passed in zero subject = 27 + 65 + 90 + 15 + 28 + 22 + 10 = 257
(xii) Number of candidates passed in at most one subject = candidates passed in one subject + candidates passed in zero subject = 15 + 28 + 22 + 10 = 75