CARDINALITY OF SETS

About the topic "Cardinality of sets"

Cardinality of sets :

Cardinality of a set is a measure of the number of elements in the set.

For example, let A = { -2, 0, 3, 7, 9, 11, 13 }

Here, n(A) stands for cardinality of the set A

And n (A) = 7

That is, there are 7 elements in the given set A.

In case, two or more sets are combined using operations on sets, we can find the cardinality using the formulas given below.

Formula 1 :

n(A u B) = n(A) + n(B) - n(A n B)

If A and B are disjoint sets, n(A n B) = 0

Then, n(A u B) = n(A) + n(B)

Formula 2 :

n(A u B u C) = n(A) + n(B) + n(C) - n(A n B) - n(B n C) - n(A n C) + n(A n B n C)

If A, B and C are all disjoint sets,

n(A n B) = 0, n(B n C) = 0, n(A n C) = 0, n(A n B n C) = 0

Then, n(A u B) = n(A) + n(B) + n(C)

Cardinality of sets - Solving word problems

Let us know some basic stuff about the above formulas which we require to solve word problems on "Cardinality of sets".

n(AuB) = Total number of elements related to any of the two events A & B.

n(AuBuC) = Total number of elements related to any of the three events A, B & C.

n(A) = Total number of elements related to A.

n(B) = Total number of elements related to B.

n(C) = Total number of elements related to C.

For three events A, B & C, we have

n(A) - [n(AnB) + n(AnC) - n(AnBnC)] = Total number of elements related to A only.

n(B) - [n(AnB) + n(BnC) - n(AnBnC)] = Total number of elements related to B only.

n(C) - [n(BnC) + n(AnC) + n(AnBnC)] = Total number of elements related to C only.

n(AnB) = Total number of elements related to both A & B

n(AnB) - n(AnBnC) = Total number of elements related to both (A & B) only.

n(BnC) = Total number of elements related to both B & C

n(BnC) - n(AnBnC) = Total number of elements related to both (B & C) only.

n(AnC) = Total number of elements related to both A & C

n(AnC) - n(AnBnC) = Total number of elements related to both (A & C) only.

For two events A & B, we have

n(A) - n(AnB) = Total number of elements related to A only.

n(B) - n(AnB) = Total number of elements related to B only.

Let us consider the following example, to have better understanding of solving word problems on "Cardinality of sets"

Example:

In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games.

Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.

Venn diagram related to the above situation:

From the venn diagram, we can have the following details.

No. of students who play foot ball = 65

No. of students who play foot ball only = 28

No. of students who play hockey = 45

No. of students who play hockey only = 18

No. of students who play cricket = 42

No. of students who play cricket only = 10

No. of students who play both foot ball & hockey = 20

No. of students who play both (foot ball & hockey) only = 12

No. of students who play both hockey & cricket = 15

No. of students who play both (hockey & cricket) only = 7

No. of students who play both foot ball and cricket = 25

No. of students who play both (foot ball and cricket) only = 17

No. of students who play all the three games = 8

Cardinality of sets - Practice problems

Let us look at some practice problems on "Cardinality of sets"

Problem 1 :

In a survey of university students, 64 had taken mathematics course, 94 had taken chemistry course, 58 had taken physics course, 28 had taken mathematics and physics, 26 had taken mathematics and chemistry, 22 had taken chemistry and physics course, and 14 had taken all the three courses. Find how many had taken one course only.

Solution :

Step 1 :

Let M, C, P represent sets of students who had taken mathematics, chemistry and physics respectively

Step 2 :

From the given information, we have

n(M) = 64 , n(C) = 94, n(P) = 58,

n(MnP) = 28, n(MnC) = 26, n(CnP) = 22

n(MnCnP) = 14

Step 3 :

From the basic stuff, we have

No. of students who had taken only Math

= n(M) - [n(MnP) + n(MnC) - n(MnCnP)]

= 64 - [28+26-14]

= 64 - 40

= 24

Step 4 :

No. of students who had taken only Chemistry

= n(C) - [n(MnC) + n(CnP) - n(MnCnP)]

= 94 - [26+22-14]

= 94 - 34

= 60

Step 5 :

No. of students who had taken only Physics

= n(P) - [n(MnP) + n(CnP) - n(MnCnP)]

= 58 - [28+22-14]

= 58 - 36

= 22

Step 6 :

Total no. of students who had taken only one course

= 24 + 60 + 22

= 106

Hence, the total number of students who had taken only one course is 106

Alternative Method (Using venn diagram)

Step 1 :

Venn diagram related to the information given in the question:

Step 2 :

From the venn diagram above, we have

No. of students who had taken only math = 24

No. of students who had taken only chemistry = 60

No. of students who had taken only physics = 22

Step 3 :

Total no. of students who had taken only one course

= 24 + 60 + 22

= 106

Hence, the total number of students who had taken only one course is 106

Let us look at the next problem on "Cardinality of sets"

Problem 2 :

In a group of students, 65 play foot ball, 45 play hockey, 42 play cricket, 20 play foot ball and hockey, 25 play foot ball and cricket, 15 play hockey and cricket and 8 play all the three games. Find the total number of students in the group.
(Assume that each student in the group plays at least one game.)

Solution :

Step 1 :

Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively.

Step 2 :

From the given information, we have

n(F) = 65 , n(H) = 45, n(C) = 42,

n(FnH) = 20, n(FnC) = 25, n(HnC) = 15

n(FnHnC) = 8

Step 3 :

From the basic stuff, we have

Total number of students in the group = n(FuHuC)

= n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC)

= 65 + 45 + 42 -20 - 25 - 15 + 8

= 100

Hence, the total number of students in the group is 100

Alternative Method (Using venn diagram)

Step 1 :

Venn diagram related to the information given in the question:

Step 2 :

Total number of students in the group

= 28 + 12 + 18 + 7 + 10 + 17 + 8

= 100

Hence, the total number of students in the group is 100

When students go through the above two example word problems on "Cardinality of sets", we hope they would have received answer for the question, "How to solve word problems on "Cardinality of sets"?". Let us see some more examples to have better understanding of solving word problems on "Cardinality of sets"

Problem 3 :

In a class of 60 students, 40 students like math, 36 like science, 24 like both the subjects. Find the number of students who like

(i) Math only, (ii) Science only (iii) Either Math or Science (iv) Neither Math nor science

Solution :

Step 1 :

Let M and S represent the set of students who like math and science respectively.

Step 2 :

From the information given in the question, we have

n(M) = 40, n(S) = 36, n(MnS) = 24

Step 3 :

Answer (i) : No. of students who like math only

= n(M) - n(MnS)

= 40 - 24

= 16

Step 4 :

Answer (ii) : No. of students who like science only

= n(S) - n(MnS)

= 36 - 24

= 12

Step 5 :

Answer (iii) : No. of students who like either math or science

= n(M or S)

= n(MuS)

= n(M) + n(S) - n(MnS)

= 40 + 36 - 24

= 52

Step 6 :

Answer (iv) :

Total no. students who like any of the two subjects = n(MuS) = 52

No. of students who like neither math nor science

= 60 - 52

= 8

Let us look at the next problem on "Cardinality of sets"

Problem 4 :

At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. Find the number of women doctors attending the conference.

Solution :

Step 1 :

Let M and D represent the set of Indian men and Doctors respectively.

Step 2 :

From the information given in the question, we have

n(M) = 23, n(D) = 4, n(MuD) = 24,

Step 3 :

From the basic stuff, we have

n(MuD) = n(M) + n(D) - n(MnD)

24 = 23 + 4 - n(MnD)

n(MnD) = 3

n(Indian Men and Doctors) = 3