Question

If n(A)=2,n(B)=m and the number of relations from A to B is 64 ,then the value of m is

Answer 1

no. of relations from set A to set B =2^n(A)n(B)
=2^2m
2^2m=64
2^2m=2^(2)(3)
hence m= 3

Answer 2

The value of m = 3

Given :

n(A) = 2 , n(B) = m and the number of relations from A to B is 64

To find :

The value of m

Concept :

1. Relation :

Let S and T be two non empty sets. A binary relation R between S and T is a subset of S × T. If the ordered pair (s, t) ∈ R then the element s of the set S is said to be related to t of the set T by the relation R

2. The number of relations from a set with m elements to a set with n elements

$\sf = {2}^{mn}$

Solution :

Step 1 of 2 :

Write down number of elements in A and B

Here it is given that n(A) = 2 , n(B) = m

Number of elements in A = n(A) = 2

Number of elements in B = n(B) = m

Step 2 of 2 :

Find the value of m

We know that the number of relations from a set with m elements to a set with n elements

$\sf = {2}^{mn}$

Since n(A) = 2 , n(B) = m

So number of relations from A to B

$\displaystyle \sf{ = {2}^{(2 \times m)} }$

$\displaystyle \sf{ = {2}^{(2 m)} }$

By the given condition

$\displaystyle \sf{ {2}^{(2 m)} = 64}$

$\displaystyle \sf{ \implies {2}^{(2 m)} = {2}^{6} }$

$\displaystyle \sf{ \implies 2m = 6}$

$\displaystyle \sf{ \implies m = 3}$

Hence the value of m = 3

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ write the following sets in roster form

N={x:x is odd number, 21<x<35,XEN

https://brainly.in/question/24185793

2. The number of non trivial relations on the set A=(a,b,c) is

https://brainly.in/question/24645961

#SPJ3