# Cardinal Number of a Finite and Infinite Set

The number of distinct elements in a finite set A is called cardinal number and it is denoted by n(A). And if it is not finite set, then it is called infinite set.

e.g., If A = {-3, -1, 8, 10, 13, 17},

then n(A) = 6

**Properties **

If A, B and C are finite sets andU be the finite universal set, then

*n*(A $\cup$ B) =*n*(A) +*n*(B) -*n*(A $\cap$ B)*If A B and are disjoint sets then, n*(A $\cup$ B) =*n*(A) +*n*(B)*n*(A $\cup$ B $\cup$ C) =*n*(A) +*n*(B) +*n*(C) -*n*(A $\cap$ B) -*n*(B $\cap$ C) +*n*(A $\cap$ B $\cap$ C)*n*(A - B) =*n*(A) -*n*(A $\cap$ B)*n*(A $\Delta$ B) =*n*(A) +*n*(B) - 2*n*(A $\cap$ B)*n(*A') =*n*(U) -*n*(A)*n*(A' $\cup$ B') =*n*(U) -*n*(A $\cap$ B)*n*(A' $\cap$ B') =*n*(U) -*n*(A $\cup$ B)*n*(A $\cap$ B') =*n*(A) -*n*(A $\cap$ B)

**Method to Find Common Roots **

Sometimes the number of common elements cannot found easily y by solving the given sets. That type of problems can be solved by drawing a curves.

The intersection point of a curve is equal to the number of common elements in a set.

e.g., Consider the sets

A = {(x,y)| y = $\frac{1}{x}$, 0 $\neq$ x $\in$ R}

and A = {(x,y)| y = -x, x $\in$ R}

the determine *n*(A $\cap$ B).

Here, we see that, ∀ x, we get infinite values of y. Hence, we find infinite sets A B and . And it is difficult to find the common elements between A and B.

Now, firstly we make the graph of given sets.

A = The set of all points on a curve xy = 1,

[xy = *c* is a rectangular hyperbola curve]
and B = The set of all points on a curve y = - x.

[y = -x is a straight which passes through origin]

Since, there is no intersection point on a curve. So, there is no common elements between two sets.**Sample Problem 1**

If A and B are sets such that n(A) = 9, n(B) = 16 and n(A $\cup$ B) = 25, find A $\cap$ B.

Answer:

We have, *n*(A $\cup$ B) = *n*(A) + *n*(B) - *n*(A $\cap$ B)

Therefore, substituting the values we get

25 = 9 + (16) - *n*(A $\cap$ B)

*n*(A $\cap$ B) = 0

So, A $\cap$ B = 0

**Sample Problem 2**

If A and B are sets such that n(A) = 14, n(A $\cup$ B) = 26 and n(A $\cap$ B) = 8, then find n(B).

Answer:

We have, *n*(A $\cup$ B) = *n*(A) + *n*(B) - *n*(A $\cap$ B)

*n*(B) = *n*(A $\cup$ B) + *n*(A $\cap$ B) - *n*(A)

*n*(B) = 26 + 8 - 14 = 20

**Sample Problem 3**

In a town of 10000 families it was
found that 40% families buy newspaper A, 20% families buy
newspaper B and 10% families buy newspaper C, 5% buy
A and B, 3% buy B and C and 4% buy A and C. If 2% families
buy all of three newspapers, then the number of families which
buy A only, is

(a) 4400

(b) 3300

(c) 2000

(d) 500

Answer: (b)

*n*(A) % = 40 of 10000 = 4000, *n*(B) = 2000, *n*(C) =1000, *n*(A $\cap$ B) = 500, *n*(B $\cap$ C) = 300, *n*(C $\cap$ A) = 400, and *n*(A $\cap$ B $\cap$ C) = 200

so, *n*(A $\cap$ $\bar{B}$ $\cap$ $\bar{C}$)

= *n*{A $\cap$ (B $\cup$ C)}

= *n*(A) - *n*{A $\cap$ (B $\cup$ C)}

= *n*(A) - *n*(A $\cap$ B) - *n*(A $\cap$ C) + *n*(A $\cap$ B $\cap$ C)

= 4000 - 500 - 400 + 200 = 3300

**Sample Problem 4**

In a survey of 600 students in a school,
150 students were found to be taking tea and 225 taking coffee.
100 were taking both tea and coffee. Find how many students
were taking neither tea nor coffee? [NCERT]

(a) 310

(b) 320

(c) 327

(d) 325

Answer: (d)

Let C and T denote the students taking coffee and tea, respectively.

Here, n(T) = 150, n(C) = 225, n(C $\cap$ T) = 100

Using the identity n(C $\cup$ T) = n(T) + n(C) - n(C $\cap$ T)

We have,

n(C $\cup$ T) = 150 + 225 - 100 = 275

Given, total number of students = 600 = n(U)

We are to find the number of students taking neither tea nor coffee i.e., n(C $\cup$ T)'.

So, n(C $\cup$ T)' = n(U) - n(C $\cup$ T) = 600 - 275 = 325

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