Test: Counting Theorems - Commerce MCQ

n (P) = 5, n(Q) = 12 and n(PUQ) = 14
n(PUQ) = n(P) + n(Q) - n(P∩Q) 
14 = 5 + 12 - n(P∩Q)
n(P∩Q) = 3

As a and b are disjoint sets, they have no elements in common. Thus,
n(a∪b) = n(a) + n(b) 
→n(b) = n(a∪b) − n(a)
= 475 − 435
= 40

A and B are disjoint set so their is no common element between A and B.
n(AuBuC) = n(A)+n(B)+ n(C)-n(AnB)-n(Anc)-n(BnC)+n(AnBnC)
As we know n(AuB)=n(A)+ n(B)-n(AnB)
Here when we add A and B we are adding (AnB) two times , as it is present in both A and B. So we subtract (AnB) one time from A and B.
Which means we need to take the area of A and B set but no area should repeat.
When we take three sets A,B and C
We take area of A+ Area of B + area of C ————eq
But they are intersecting so the area we are taking is greater than actual area. We have again added (AnB), (AnC) and (BnC) twice. Therefore we subtract the intersecting area from eq one time as done above
A+B+C-(AnB)-(AnC)-(BnC)
 
The intersection of the sets a and b is the set of all the elements which belong to both A and B. It is denoted by A ∩ B (“A intersection B”).
• If A and B do not have any element in common then A ∩ B= a null set = Ø.

r :- c
Explanation:- Students got distinction in mathematics = 24
 Student got distinction in physics = 19 
Students got distinction in both= 15
Total number of students = (24+19)-15
= 28

Let S = {Persons who drive scooter}
C = {Persons who drive car}
Given, n(S ∪ C) = 80, n(S) = 35, n(C) = 60
Therefore, n(S ∪ C) = n(S) + n(C) - n(S ∩ C)
80 = 35 + 60 - n(S ∩ C)
80 = 95 - n(S ∩ C)
Therefore, n(S ∩ C) = 95 – 80 = 15
Therefore, 15 persons drive both
scooter and car.
Therefore, the number of persons who
drive a scooter only = n(S) - n(S ∩ C)
= 35 – 15
= 20
Also, the number of persons who drive car only = n(C) - n(S ∩ C)
= 60 - 15
= 45

n(P ∪ Q) = n(P)  +  n(Q) - n(P ∩ Q) 
140 = 120 + 50 - x
140 = 170 -x
x = 30

Correct option is D.
To find A×B we take one element from set A and one from set B. 

Given that 99 elements are common to both set A and set B.

Suppose these common elements are N_1​,N₂​,N₃​,...N₉₉​. 

Select an ordered pair for A×B such that both are selected out of these common elements. Examples: (N₁​,N_2​),(N₃​,N_5​)
All these will also be elements of B×A. Hence number of elements common to A×B and B×A is 99×99=99² ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)

n[(A×B)∩(B×A)]=n[(A∩B)∩(B∩A)]=(99)(99)=99²

By demorgan's law
n( A' ∩ B' )=n(A ∪ B)' = n(U) - n(A ∪ B)
=100 - [ n(A) + n(B) - n(A ∩ B) ]
=100 - 60 = 40

The number of non- empty subsets =
2ⁿ − 1 = 2⁴ − 1 = 16 − 1 = 15

P(U) = 120
Mathematics = P(A) = 90
Economics = P(B) = 72
P(A ∪ B)' = 10
P(A ∪ B) = P(U) - P(A ∪ B)'
= 120 - 10 = 110
P(A ∪ B) = P(A) + P(B) - P(A ∪ B)
110 = 90 + 72 -P(A ∩ B)
P( A ∩ B) = 162 - 110 = 52

Total students who took any of the subjects = 40 - x
Let x be the no. of students who took neither of the subjects
Then, 40 - x = 14 + 29 - 5
⇒ x = 2

N(A) indicates the number of elements in set A, there are 4 elements.

A={4,6,10} B={6,8}
A-B={4,10}
= n(A-B) = 2 Elements

Here A ∩ C = Ø ⇒ A ∩ B ∩ C = Ø
n (A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)
 = 25 + 20 + 27 - 5 - 7 - 0 - 0
= 60

let the total number of workers be 100. A = who likes oranges , B = who likes apples
So n(A ) = 70 and n(B) = 64, n(A ∩ B) = x
Also n(A U B) ≤ 100
n(A) + n(B) -n(A ∩ B) ≤ 100
70 + 64 - x ≤ 100
x ≥ 34
and n(A ∩ B) ≤ n(B) ⇒ x ≤ 64
then 34 ≤ x ≤ 64