Mess with set theory [closed]

There are 3 sets $A$, $B$ and $C$ included in $U$ such that:

\begin{cases} C\cap A =C & \\ n(C)^{c}=150 \\ n(A^{c}\cap B^{c} ) ^{c}=90\\ n[(A\cup B)-C] =6n(C)\\ n(U)=? \end{cases}

Help with set theory problem

edit: $n$ = cardinality n(U) = Number of elements in the universe

closed as off-topic by B. Mehta, Graham Kemp, Moishe Kohan, Andrés E. Caicedo, I am BackJun 11 '17 at 2:48

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• What does $n$ denote? – B. Mehta Jun 11 '17 at 1:23
• What the heck is $n(\cup)$??? – Brethlosze Jun 11 '17 at 1:25
• I would guess it's meant to be $n(U)$. – pjs36 Jun 11 '17 at 1:26
• @Susana So, Susana, what have you tried and where is this problem giving you trouble? You get better response if you include your own work. – Graham Kemp Jun 11 '17 at 1:27
• If I knew how to start I would do it – zeros Jun 11 '17 at 1:33

I'll only provide a sketch and leave you to fill in the details.

First here's a handy counting formula that I always show my finite math students. For two sets $S$ and $T$,

$$|S \setminus T | = |S| - |S \cap T|$$

where I write $S \setminus T$ instead of $S - T$ to represent the collection of objects that are in $S$ but not in $T$.

You should also know that if $S \cap T = S$, then $S \subseteq T$, and that $(S^C \cap T^C)^C = S \cup T$, by de Morgan's laws.

Now let's get to counting. Condition 2 says that

$$|U| - |C| = 150$$

Condition 3 says $$|A \cup B| = |A| + |B| - |A \cap B| = 90$$

or that

$$|(A^C \cap B^C)^C| = |U| - |A \cup B| = 90$$

Finally condition 4 says

\begin{align*} |(A \cup B) \setminus C| &= |A \cup B| - |(A \cup B) \cap C| \\ &= |A \cup B| - |(A \cap C) \cup (B \cap C)| \\ &= 6|C| \end{align*}

where I have used the distributive property of intersection over union. Now it's just up to some clever algebra.

• I do not clarify with the venn diagram, help please – zeros Jun 11 '17 at 23:30

Hint: Draw a Venn diagram of the sets.