Question

Solve the question no. 24.

Answer 1

Question:

Two finite sets have m and n elements. The number of elements in power set of first set is 48 more than the total number of elements in power set of the second set. Then the values of m and n are :

a). 7,6

b). 6,3

c). 6,4

d). 7,4

Answer:

c). 6,4

Note:

• Set : A well defined collection of distinct objects is called set .

• Number of elements in a finite set A (cardinal number) is denoted by n(A) .

• Power set : The set or a family of all the subsets of a given set A is said to be the power set of a and is denoted by P(A) .

• If a finite set A has n elements, then its power set has 2^k elements , ie ; n[P(A)] = 2^k

Solution:

Here,

It is given that;

The number of elements in first set is n .

Thus, the number of elements in the power set of first set will be 2^n .

Also,

The number of elements in second set is m .

Thus, the number of elements in the power set of second set will be 2^m .

Now,

According to the question;

The number of elements in power set of first set is 48 more than the total number of elements in power set of the second set.

Thus,

=> 2^m = 2^n + 48

=> 2^m - 2^n = 48

=> (2^n)•[2^(m-n) - 1] = (2^4)•3

=> (2^n)•[2^(m-n) - 1] = (2^4)•[4 - 1]

=> (2^n)•[2^(m-n) - 1] = (2^4)•[2^2 - 1]

Now,

Comparing both the sides of above equation,

We have;

n = 4 and (m-n) = 2

Thus,

=> m - n = 2

=> m - 4 = 2

=> m = 2 + 4

=> m = 6

Hence,

The required values of m and n are 6 and 4 respectively .

Answer 2

Question :- Two finite sets have m and n elements. The number of elements in power set of first set is 48 more than the total number of elements in power set of the second set. Then the values of m and n will be ?

Concept used :--

• If a finite sets have n elements , than number of subsets will be 2^n .
• a^m - a^n = a^(m-n)

Solution :---

The no. Of element in 1st set = m

No. Of subsets of 1st set = 2^m

No. Of element ts in 2nd set = n

No. Of subsets of 2 nd set = 2^ n

According to question now,,

2^m - 2^n = 48

Taking 2^n common From LHS

2^n{(2^m/2^n) - 1}=16×3

2^n{2^(m - n )- 1}= 2⁴×{(2²-1}

On comparing we get,,

→ n = 4

→ (m-n) = 2

or,

→ m - 4 = 2

→ m = 6

Hence, the value of m and n will be 6 and 4 . Option (C) .