Question

what is the formula of n(AnBnC)​

Answer 1

Answer:

their is no formula for intersecting bro ...

n(AnBnC) is the simply values of all common part in Venn diagram ...

Answer 2

n(AnBnC) = n(AuBuC) - n(A) - n(B) - n(C) + n(AnB) + n(Anc) + n(BnC)

Derivation:

We derive the formula of n(AnBnC) using the formula;

n(AuBuC) = n(A)+n(B)+ n(C)-n(AnB)-n(Anc)-n(BnC)+n(AnBnC)

WKT;

n(AuB) = n(A)+ n(B)-n(AnB)

• Here, when we are adding A and B we are adding (AnB) two times , as it is present in both A and B. So subtract the (AnB) one time from A & B.

• This means we need to take the area of A and B set but no area should repeat.

Similarly,

For three sets A,B and C;

We take (Area of A+ Area of B + Area of C)

• They are intersecting so area we are taking is greater than the actual area.
• We have added (AnB), (AnC) &(BnC) twice. Therefore we subtract the intersecting area from eq one time as done below.

                                      A+B+C-(AnB)-(AnC)-(BnC)

• But when we are doing this, area of (AnBnC) is also removed as A,B,C,(Anb),(Anc),(BnC) all of them contain (AnBnC) in them.

Therefore we add a (AnBnC) in equation.

n(AuBuC) = n(A)+n(B)+ n(C)-n(AnB)-n(Anc)-n(BnC)+n(AnBnC)

=>

n(AnBnC) = n(AuBuC) - n(A) - n(B) - n(C) + n(AnB) + n(Anc) + n(BnC)

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