Assume $ A \subset B$ is true. 

Then, every element of $A$ i.e. $a _1,a _2, ... , a _n$ in A are also in B.

So, number of elements in $B$ will always be greater than no. of elements in $A$

And $P(A)$ will contain less number of subsets than $P(B)$

Hence, $n[P(A)] <  n[P(B)]$

$\displaystyle n\left( A\quad \cup \quad B \right) =n\left( A \right) +n\left( B \right) -n\left( A\quad \cap \quad B \right) $

If, $n(A)=n(B)$

a.  $n(A-B) = n(B-A)$. As, the no. of elements are same, if we subtract A from B or B from A, we will get the same no. of elements

b.  $n(AB)\neq n(A)+n(B)$. It is $n(A\cup B)=n(A)+n(B)$

c.  $n(A-B)= \phi$

d.  $n(AB)\neq n(B) - n(A-B)$. It is $n(AB) = n^2$ where n is the no. of elements of these sets

$$n\left( A \right) =5$ 

If there is a finite number of n elements in $A,$ then the power set $P (A)$ has $2^n$ elements. 

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$=200+300-100$
$=400$
$n(A'\cap B')=n(A\cup B)'$
                    $=n(U) - n(A\cup B)$
$300=n(U)-400$
$n(U)=700$

Using De Morgan's law,
$\displaystyle n(A'\cup B')= n(A\cap  B)' $ $\displaystyle

= n(U)-n(A\cap B)=  n(U)-100 $ $\displaystyle = 700-100= 600$

$n(A)  =1620, n(B) = 1500$

Let $A=$ number of people like coffee, $B=$ number of people like tea.

$\displaystyle n \left ( A' \cap B'\right )=n\left ( A\cup  B\right

)'$ $\displaystyle =n\left ( u \right )-n\left ( A\cup B \right

)$ $\displaystyle =n\left ( u \right )-\left {n \left ( A \right

)+n\left ( B \right )-n\left ( A\cap B \right ) \right

}$ $\displaystyle =700-\left { 200+300-100 \right }=300$

Given $\displaystyle n\left( A \right) =70$ and $\displaystyle n\left( B \right) =60$ and $\displaystyle n\left( A\quad \cup \quad B \right) =110 $.

We know,  $n(A\cup B)=n(A)+n(B)-n(A\cap B)$

In all cases(even if sets A and B are disjoint)

The square of every integer is a non-negative number
$\displaystyle \therefore A = \phi   $ i.e. the null set.

Total possible numbers of form $\dfrac { p }{ q } $ when $p\neq q$ is $={}^{ 6 }C _{ 2 }=30$

Numbers when $p=q$ is $={}^{ 6 }C _{ 1 }=6$

Therefore total numbers $30+6=36$

$\dfrac { 1 }{ 1 } =\dfrac { 2 }{ 2 } =\dfrac { 3 }{ 3 } =\dfrac { 4 }{ 4 } =\dfrac { 5 }{ 5 }= \dfrac { 6 }{ 6 } $      (five numbers deducted from caardinality of set) 

$\dfrac { 1 }{ 2 } =\dfrac { 2 }{ 4 } =\dfrac { 3 }{ 6 } $    (two numbers deducted from caardinality of set) 

$\dfrac { 2 }{ 1 } =\dfrac { 4 }{ 2 } =\dfrac { 6 }{ 3 } $     (two more  numbers deducted from caardinality of set) 

$\dfrac { 1 }{ 3 } =\dfrac { 2 }{ 6 } $          (one number deducted from caardinality of set) 

$\dfrac { 3 }{ 1 } =\dfrac { 6 }{ 2 } $          (one more number deducted from caardinality of set) 

$\dfrac { 2 }{ 3 } =\dfrac { 4 }{ 6 } $          (one number deducted from caardinality of set) 

$\dfrac { 3 }{ 2 } =\dfrac { 6 }{ 4 } $          (one more number deducted from caardinality of set) 

So, the caardiality of set $=36-5-2-2-1-1-1-1=23$.

So, option B is correct.

If $A$ is a set. Then, $n(A)$ is called cardinal number of the set A = number of elements in set A.

$P(A)$ is called power set of $A$. Power set $= P(A) =$ set of all subsets of set $A$

We know that if number of elements in set $A$ is $n$

$n(A) = n$

Number of subsets of power set $= 2 ^n$

Given,

$n[ P(A)] = 16$

$2^n = 16$

$2 ^n = 2 ^4$

Therefore, $n = 4$

In A, last term will be $400$.

Let $F,H$ and $C$ denote the no. of children who play Football, Hockey and Cricket respectively.
Given $n(F)=35,n(H)=22,n(C)=25$ and $n(F\cap C\cap H^{\prime})=7,n(F\cap H\cap C^{\prime})=3,n(F\cap C)=12$
but $n(F\cap C\cap H^{\prime})=n(F\cap C)-n(F\cap C\cap H)$
$\Rightarrow 7=12-n(F\cap C\cap H)$
$\Rightarrow n(F\cap C\cap H)=5$
$\therefore$ No of children who play all three games is $5$
Hence, option A