If A and B are two sets having 3 elements in common. If n(A) = 5, n(B)

1.	If A and B are two sets having 3 elements in common. If n(A) = 5, n(B) = 4, find n(A x B) and n[(A x B)∩ (B x A)].
Answer» Given: (A) = 5 and n(B) = 4 To find: [(A × B) ∩ (B×A)] n (A × B) = n(A) × n(B) = 5 x 4 = 20 n (A ∩ B) = 3 (given: A and B has 3 elements in common) In order to calculate n [(A × B) ∩ (B × A)], We will assume, A = (x, x, x, y, z) and B = (x, x, x, p) So, we have (A × B) = {(x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (y, x), (y, x), (y, x), (y, p), (z, x), (z, x), (z, x), (z, p)} (B × A) = {(x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (p, x), (p, x), (p, x), (p, y), (p, z)} [(A × B) ∩ (B × A)] = {(x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x)} ∴ We can say that n [(A × B) ∩ (B × A)] = 9

If A and B are two sets having 3 elements in common. If n(A) = 5, n(B) = 4, find n(A x B) and n[(A x B)∩ (B x A)].

Answer»

Given: (A) = 5 and n(B) = 4

To find: [(A × B) ∩ (B×A)]

n (A × B) = n(A) × n(B) = 5 x 4 = 20

n (A ∩ B) = 3

(given: A and B has 3 elements in common)

In order to calculate n [(A × B) ∩ (B × A)],

We will assume,

A = (x, x, x, y, z) and B = (x, x, x, p)

So, we have

(A × B) = {(x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (y, x), (y, x), (y, x), (y, p), (z, x), (z, x), (z, x), (z, p)}

(B × A) = {(x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (p, x), (p, x), (p, x), (p, y), (p, z)}

[(A × B) ∩ (B × A)] = {(x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x)}

∴ We can say that n [(A × B) ∩ (B × A)] = 9

Discussion

No Comment Found

Related InterviewSolutions

In fig. shows a relationship between the sets P and Q. Write this relation in :a. Set builder form b. Roster form c. What is its domain and range?
Let A = {1, 2, 3, 4, 5, 6}. Let R be a relation on A defined by R = {(a, b): a, b A, b is exactly divisible by a} i. Write R in roster form ii. Find the domain of R iii. Find the range of R.
Let `A = {1, 2, 3, 4, 6}`. Let R be the relation on A defined by `{(adot b): a , b in A , b`is exactly divisible by a}.(i) Write R in roster form(ii) Find the domain of R(iii) Find the range of R.
Let `A={1,2,3,5}" and "B={4,6,9}.` Define a relation from A to B, given by `R={(a,b):a inA,b inB" and "(a-b)" is odd"}.` (i) Write R in roster form. (ii) Find dom (R) and range (R).
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides a – b}, is an equivalence relation.
Let W denote the words in the English dictionary. Let the relation R be defined by R = {(x, y) ∈ W × W : the words x and y have at least one letter in common}. Then R is (a) Reflexive and transitive, not symmetric (b) Reflexive and symmetric and not transitive (c) Symmetric and transitive, not reflexive (d) Reflexive, symmetric and transitive
Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which can be added to R to make it an equivalence relation is(a) 2 (b) 3 (c) 5 (d) 7
If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric
Let I be the set of integers and R be a relation on I defined by R = {(x, y) : (x – y) is divisible by 11, x, y ∈ I}. Then R is(a) An equivalence relation (b) Symmetric only (c) Reflexive only (d) Transitive only
Let Z be the set of integers. Then the relation R = {(a, b) : a, b ∈ Z and (a + b) is even} defined on Z is(a) Only symmetric (b) Symmetric and transitive only (c) An equivalence relation (d) None of the above