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) - 2n(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]
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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