Question

Let $A,B$ and $C$be finite sets. Suppose that $n(A)=10,n(B)=15,n(C)=20,n(A\cap B)=8$ and $n(B\cap C)=9$. Then the possible values of $n(A\cup B\cup C)$ is
A. \[26\]
B. \[27\]
C. \[28\]
D. Any of the three values \[26\], \[27\] and \[28\] is possible.

Answer 1

Hint: To solve this question, we will substitute the given values in the formula $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$ and form an equation for $n(A\cup B\cup C)$using the relation $n(A\cap C)\ge n(A\cap B\cap C)$. Then we will calculate the value of $n(A\cup B)$and $n(B\cup C)$ using formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$. Using values of $n(A\cup B)$ and $n(B\cup C)$ we will form an equation for $n(A\cup B\cup C)$. Using both the equations of $n(A\cup B\cup C)$ we will determine the possible values of \[26\], \[27\] and \[28\].



Formula Used:$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$



Complete step by step solution:We are given three finite sets $A,B$ and $C$ with values $n(A)=10,n(B)=15,n(C)=20,n(A\cap B)=8$ and $n(B\cap C)=9$ and we have to find the possible values of $n(A\cup B\cup C)$.
We will first substitute the given values in the formula.
$\begin{align}
  & n(A\cup B\cup C)=10+15+20-8-9-n(A\cap C)+n(A\cap B\cap C) \\
 & =28-[n(A\cap C)-n(A\cap B\cap C)]......(i)
\end{align}$
As we know that $n(A\cap C)\ge n(A\cap B\cap C)$so we can write it as $n(A\cap C)-n(A\cap B\cap C)\ge 0....(ii)$.
From equation (i) and (ii) we can say that,
$\begin{align}
  & 28-n(A\cap C)+n(A\cap B\cap C)\ge 0 \\
 & n(A\cap B\cap C)\le 28....(iii)
\end{align}$
We will calculate the value of $n(A\cup B)$.
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
$\begin{align}
  & n(A\cup B)=10+15-8 \\
 & =17.....(iv)
\end{align}$
We will now calculate the value of $n(B\cup C)$.
 $\begin{align}
  & n(B\cup C)=n(B)+n(C)-n(B\cap C) \\
 & =15+20-9 \\
 & =26....(v)
\end{align}$
From equation (iv) and (v),
We can write $n(A\cup B\cup C)\ge 26....(vi)$ and $n(A\cup B\cup C)\ge 17...(vii)$.
From equation (ii), (vi) and (vii) we can say that,
$26\le n(A\cup B\cup C)\ge 28$
Hence we can say that the possible value of $n(A\cup B\cup C)$ will be either \[26\], \[27\] or \[28\].



Option ‘D’ is correct



Note: We have used the relation $n(A\cap C)\ge n(A\cap B\cap C)$ in the solution. The value of $n(A\cap C)$will be always greater than or equal to $n(A\cap B\cap C)$because $n(A\cap C)$ is the intersection of only two sets $A$ and $C$ which means that it will contain only those elements that are common in both sets while $n(A\cap B\cap C)$ will contain all the elements which will be common in all three sets $A,B$ and $C$.