Question

If X and Y are two sets such that $n(X) = 17,n(Y) = 23$ and $n(X \cup Y) = 38$, find $n(X \cap Y)$.

Answer 1

Two sets $A$ and $B$ are said to be disjoint, if $A \cap B = \phi$. If $A \cap B \ne \phi$, then $A$ and $B$ are called overlapping or intersecting sets.
$n$, the number of elements in a set is called the cardinal number of the set, denoted by $n(A)$.
Cardinal Property of Sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$ and if $A$ and $B$ are disjoint, then $n(A \cup B) = n(A) + n(B)$.

Complete step by step answer:
We know that any well-defined collection of objects, which are different from each other, and which we can see or think of is called a set. The objects which belong to a set are called its members or elements.
Two sets are said to be disjoint if they have no element in common.
We are given two sets $X$ and $Y$ such that $n(X) = 17,n(Y) = 23$and$n(X \cup Y) = 38$.
We have to find the value of $n(X \cap Y)$.
According to the Cardinal property of sets we have,
If $X$ and $Y$ are two joint sets then $n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$.
By substituting the values given in the question, we get
$\Rightarrow 38 = 17 + 23 - n(X \cap Y)$
By transposing $n(X \cap Y)$ to the L.H.S and $38$ to the R.H.S we get,
$\Rightarrow n(X \cap Y) = 17 + 23 - 38$
On solving the above equation
$\Rightarrow n(X \cap Y) = 40 - 38$
$\Rightarrow n(X \cap Y) = 2$

$\therefore$ The value of $n(X \cap Y) = 2$.

Note:
The union of two sets $A$ and $B$, denoted by $A \cup B$ (read as $A$ union $B$), is the set of all those elements which are either in sets $A$ or in sets $B$ or in both.
Symbolically, $A \cup B = \\{ x:x \in A \, or \, x \in B\\}.$
The intersection of two sets $A$ and $B$, denoted by $A \cap B$ (read as $A$ intersection $B$), is the set of all those elements which are common to both sets $A$ and $B$.
Symbolically, $A \cap B = \\{ x:x \in A\, and\, x \in B\\}.$
Union and Intersection are two main set operations.