Tag: cardinal number of a finite set

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Questions Related to cardinal number of a finite set

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $A\subset B$, then $n[P(A)]$ ______ $n[P(B)]$

  1. $=$
  2. $<$
  3. $\leq $
  4. $>$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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, $n[P(A)] <  n[P(B)]$

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $A $and $B$ are not disjoint, then $\displaystyle n\left( A \cup  B \right) $ is equal to

  1. $\displaystyle n\left( A \right) +n\left( B \right) $
  2. $\displaystyle n\left( A \right) +n\left( B \right) -n\left( A \cap B \right) $
  3. $\displaystyle n\left( A \right) +n\left( B \right) +n\left( A \cap B \right) $
  4. $\displaystyle n\left( A \right) .n\left( B \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\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) $

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

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

  1. $n(A - B) = n(B - A)$
  2. $n(AB) = n(A) + n(B)$
  3. $n(A - B) =\phi$
  4. $n(AB) = n(B) - n(A - B)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

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

Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

If $n(A) = n(B)$ then:

  1. $n(A- B) = n(B- A)$
  2. $n(AB)= n(A) + n(B)$
  3. $n(A- B)=n(A)-n(B)$
  4. $n(AB) = n(B) - n(A-B)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given: $n(A) = n(B)$
$n(A)$ is the cardinal no. of set $A$ and same for the set $B$

Thus, the number of elements are always same no matter what type of operation we are performing.

Hence, $n(A-B)=n(B-A)$.
Multiple choice maths introduction to set cardinal number of a finite set cardinality of a set representation of sets

Let $U$ be the universal set for sets $A$ and $B$ such that $n(A)=200 , n(B)=300$ and $n(A\cap B)=100$, then $n(A'\cap B')$ is equal to $300$ provided that $n(U)$ is equal to

  1. $600$
  2. $700$
  3. $800$
  4. $900$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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