Question

Q: For any three sets A,B,C prove that: n(A U B U C) = [n(A) + n(B) + n(C) + n(A n B n C)] - [n(AnB) + n(BnC) + n(AnC)] Ma'am/Sir pls explain this formula to me.

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

Let A, B and C be three sets such that

$n(A \cup B \cup C)=n[A \cup(B \cup C)] \rightarrow(i)$

We know that,  

$n(A \cup B)=n(A)+n(B)-n(A \cup B)$

Now, by applying the above property in equation (i), we get

$\begin{aligned} n(A \cup B \cup C)=& n(A)+[n(B)+n(c)-n(B \cup C)] \\ &-n[(A \cap B) \cup(A \cap C)] \end{aligned}$

By applying the distributive property $A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$, we get

$\begin{array}{c}{=n(A)+n(B)+n(c)-n(B \cap C)} \\ {-[n(A \cap B)+n(A \cap C)]-n(A \cap B) \cap(A \cap C)}\end{array}$

$\begin{aligned}=n(A)+& n(B)+n(c)-n(B \cap C)-n[(A \cap B\\ &+n(A \cap C)-n(A \cap B \cap C) ] \end{aligned}$

$\begin{aligned}=& n(A)+n(B)+n(c)-n(B \cap C)-n(A \cap B) \\ &-n(A \cap C)+n(A \cap B \cap C) \end{aligned}$

Now, by rearranging the above equation, we get

$\begin{aligned} n(A \cup B \cup C)=& n(A)+n(B)+n(c)+n(A \cap B \cap C) \\ &-n(B \cap C)-n(A \cap B)-n(A \cap C) \end{aligned}$

Therefore, we get

$\begin{aligned} n(A \cup B \cup C)=& n(A)+n(B)+n(c)+n(A \cap B \cap C) \\ &-[n(B \cap C)+n(A \cap B)+n(A \cap C)] \end{aligned}$

∴ Hence it is proved that the given i.e., $\begin{array}{l}{n(A \cup B \cup C)=n(A)+n(B)+n(c)+n(A \cap B \cap} \\ {C )-[n(B \cap C)+n(A \cap B)+n(A \cap C)]}\end{array}$ is applicable for any 3 sets of A, B, C.

The proved formula explained as, that the union (A \cup B \cup C) of three sets is equal to the sum of the total numbers of 3sets and the intersection of 3 sets $n(A)+n(B)+n(c)+n(A \cap B \cap C)$ to the subtraction of the sum of “the intersection of the three sets” in the way of two-two sets $[n(B \cap C)+n(A \cap B)+n(A \cap C)]$.