Question

Two finite sets a and b have m and n elements respectively. If the total number of subsets of a is 112 more than the total number of of B, then the value of m is

Answer 1

$\large\underline{\sf{Given- }}$

↝Two finite sets a and b have m and n elements respectively.

↝The total number of subsets of a is 112 more than the total number of of B

$\large\underline{\sf{To\:Find - }}$

↝The value of m is

$\large\underline{\sf{Solution-}}$

Given that

Two finite sets a and b have m and n elements.

Now, we know that

If set A has 'n' number of elements then ,

$\boxed{ \rm \: Number \: of \: subsets \: = \: {2}^{n}}$

Now,

Here,

Set a has 'm' elements.

So,

$\boxed{ \rm \: Number \: of \: subsets \: of \: a \: = \: {2}^{m}}$

Also,

Set b has 'n' elements,

So,

$\boxed{ \rm \: Number \: of \: subsets \: of \: b \: = \: {2}^{n}}$

According to statement,

The total number of subsets of a is 112 more than the total number of of B.

It means

$\rm :\longmapsto\: {2}^{m} - {2}^{n} = 112$

can be rewritten as

$\rm :\longmapsto\: {2}^{n}({2}^{m - n} - 1) = 112$

$\rm :\longmapsto\: {2}^{n}({2}^{m - n} - 1) = 16 \times 7$

$\rm :\longmapsto\: {2}^{n}({2}^{m - n} - 1) = 2 \times 2 \times 2 \times 2 \times 7$

$\rm :\longmapsto\: {2}^{n}({2}^{m - n} - 1) = {2}^{4} \times (8 - 1)$

$\rm :\longmapsto\: {2}^{n}({2}^{m - n} - 1) = {2}^{4} \times ( {2}^{3} - 1)$

So, on comparing, we get

$\rm :\implies\:n= 4$

and

$\rm :\longmapsto\:m - n = 3$

$\rm :\longmapsto\:m - 4 = 3$

$\rm :\longmapsto\:m = 3 + 4$

$\bf\implies \:m = 7$

Additional Information :-

$\boxed{ \rm \: {U}^{'} = \phi}$

$\boxed{ \rm \: \phi \: ' \: = \: U }$

$\boxed{ \rm \: A \: \cup \: A' \: = \: U}$

$\boxed{ \rm \: A \: \cap \: A' \: = \: \phi}$

$\boxed{ \rm \: n(A \cup \: B) = n(A) + n(B) - n(A \cap \: B)}$

$\boxed{ \rm \: n(A - B) = n(A) - n(A\cap \: B)}$

$\boxed{ \rm \: n(A \cup \: B) = n(A - B) + n(B - A) + n(A \cap \: B)}$