Proving Sets

**edriann** · July 8th 2011, 03:14 PM

Hey guys. I posted this yesterday. http://www.mathhelpforum.com/math-he...rs-184242.html. And got infractions, so it was a lesson.

But now that this homework is done. Can you now teach me how to solve this thing? I got squat yesterday, I did not pass anything.

So, now I want to know how to solve it. 'Cause the Prof. just let us submit the homework and he dismissed us after. He did not explain/teach us how to do it and what is the right answer.

Please guys, it's no homework anymore, so please help me.

**HallsofIvy** · July 8th 2011, 04:17 PM

Please tell us what you did.

**Soroban** · July 8th 2011, 04:40 PM

Hello, edriann!

$\text{Prove:}$

$n(A\cup B\cup C)$
. . . . $=\:n(A) + n(B) + n(C) - b(A\cap B) - n(A\cap C) - n(B\cap C ) + n(A\cap B\cap C)$

We know the formula for two sets:
. . $n(P \cup Q) \:=\:n(P) + n(Q) - n(P\cap Q)$

We are given: $A \cup B\cup C$

Group them into two sets: . $A \cup [B \cup C]$

Then we have:

$n(A\cup [B \cup C]) \;=\;n(A) + n(B\cup C) - n(A \cap[B \cup C])$

. . $=\;n(A) + n(B) + n(C) - n(B \cap C) - n([A\cap B]\cup[A\cap C])$

. . $=\;n(A) + n(B) + n(C) - n(B \cap C)$
. . . . $- \bigg[n(A\cap B) + n(A\cap C) - n([A\cap B] \cap[A\cap C]\bigg]$

. . $=\;n(A) + n(B) + n(C) - n(B\cap C) - n(A \cap B) - n(A\cap C) + n(A\cap B\cap C)$

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