
Given $n(\bigcup )=20$, $n(A)=12$, $n(B)=9$, $n(A\cap B)=4$, where $\bigcup $ is the universal set, $A$ and $B$ are subsets of $\bigcup $, then $n\left( {{(A\cup B)}^{c}} \right)=$
A. $17$
B. $9$
C. $11$
D. $3$
Answer
162.3k+ views
Hint: In this question, we are to find the given set operation. For this, we use the addition theorem on the sets formula. On substituting the given values, we get the required value.
Formula Used: Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: Given details are:
$n(\bigcup )=20$, $n(A)=12$, $n(B)=9$, $n(A\cap B)=4$
Where $\bigcup $ is the universal set, $A$ and $B$ are subsets of $\bigcup $ i.e., $A\subset \bigcup ;B\subset \bigcup $
From the formula we have
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
On substituting,
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& \text{ }=12+9-4 \\
& \text{ }=17 \\
\end{align}$
Since we have $A\subset \bigcup ;B\subset \bigcup $,
$\begin{align}
& n\left( {{(A\cup B)}^{c}} \right)=n(\bigcup )-n(A\cup B) \\
& \text{ }=20-17 \\
& \text{ }=3 \\
\end{align}$
Thus, the required value is $n\left( {{(A\cup B)}^{c}} \right)=3$.
Option ‘D’ is correct
Note: Here we may go wrong with the formula. Generally, we write $n({{(A\cup B)}^{c}})=1-n(A\cup B)$, but if it is mentioned that $A\subset \bigcup ;B\subset \bigcup $, we should apply the universal set as $n({{(A\cup B)}^{c}})=n(\bigcup )-n(A\cup B)$. So, we get the required value.
Formula Used: Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots . \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: Given details are:
$n(\bigcup )=20$, $n(A)=12$, $n(B)=9$, $n(A\cap B)=4$
Where $\bigcup $ is the universal set, $A$ and $B$ are subsets of $\bigcup $ i.e., $A\subset \bigcup ;B\subset \bigcup $
From the formula we have
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
On substituting,
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& \text{ }=12+9-4 \\
& \text{ }=17 \\
\end{align}$
Since we have $A\subset \bigcup ;B\subset \bigcup $,
$\begin{align}
& n\left( {{(A\cup B)}^{c}} \right)=n(\bigcup )-n(A\cup B) \\
& \text{ }=20-17 \\
& \text{ }=3 \\
\end{align}$
Thus, the required value is $n\left( {{(A\cup B)}^{c}} \right)=3$.
Option ‘D’ is correct
Note: Here we may go wrong with the formula. Generally, we write $n({{(A\cup B)}^{c}})=1-n(A\cup B)$, but if it is mentioned that $A\subset \bigcup ;B\subset \bigcup $, we should apply the universal set as $n({{(A\cup B)}^{c}})=n(\bigcup )-n(A\cup B)$. So, we get the required value.
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