How to transform a set with complementary so that I can apply Inclusion-exclusion principle

Question

We have:

• U = 150
• |M| = 90
• |E| = 75
• |F| = 80
• |M ∩ E| = 45
• |M ∩ F| = 35
• |M ∩ F ∩ E| = 10

We want:

• |E ∩ F ∩ $M^c$| = ?

I know I have to use the exclusion-inclusion principle. I have to take the last statement and transform it so that it is similar to:

• |A' ∪ B'|

But I don't know how transform it; that complementary is problematic. I have tried playing with the complementary, but I get this:

$|E ∩ F ∩ M^c| = |(E∪F∪M^c)^c|$

But that is not useful, right?

I think this properties coud be useful too:

• A∩U =A
• A∩∅=∅
• A∪U=U
• A∪∅=A

I would appreciate some tips.

Edit:

My Solution:

$U = 150 = |E ∪ F ∪ M|$

1 $|M^c ∩ F ∩ E| = |F ∩ E| - |M ∩ F ∩ E|$

$|E ∪ F ∪ M| = |E| + |F| + |M| - (|E ∩ F| + |E ∩ M| + |F ∩ M|) + |E ∩ F ∩ M|$

$150 = 75 + 80 + 90 - |E ∩ F| - 45 - 35 + 10$

$|E ∩ F| = 25$

So, back to 1:

$|M^c ∩ F ∩ E| = |F ∩ E| - |M ∩ F ∩ E| = 25 - 10 = 15 $

Answer 1

We know that $E \cap F \cap M^{c} = (E \cap F) - (E \cap F \cap M)$. Thus you need to find $E \cap F$. I think a Venn Diagram will help here since you know the size of $U$ (the universe). Or one could proceed by inclusion-exclusion. We know that $$|E \cup F \cup M| = |E|+|F|+|M|- (|E \cap F|+ |E \cap M|+ |F \cap M|) + |E \cap F \cap M|$$

Answer 2

Since you know that $M\cup E\cup F = 150$, the Inclusion-Exclusion Formula gives you \begin{align*} 150 &= |M\cup E\cup F|\\ &= |M|+|E|+|F|-|M\cap E|-|M\cap F|-|E\cap F|+|M\cap E\cap F|\\ &= 90 + 75 + 80 - 45 - 35 - |E\cap F| + 10\\ &= 175 - |E\cap F|, \end{align*} so you know that $|E\cap F|= 25$.

This gives you all the information you need: For example, to divide $M$ into its four disjoint parts ($M\cap E\cap F$, $M\cap E^c\cap F$, $M\cap E\cap F^c$, and $M\cap E^c\cap F^c$), you use inclusion-exclusion:

• You know that $M$ contains $10$ elements that are also in $E$ and in $F$ (that is, $|M\cap E\cap F| = 10$).
• $M$ contains $35$ elements that are also in $E$ but not in $F$ (because $|M\cap E| = |M\cap E\cap F^c| + |M\cap E\cap F|$, and you know that $|M\cap E| = 45$ and $|M\cap E\cap F|=10$).
• $M$ contains $25$ elements that are also in $F$ but not in $E$ (because $|M\cap E^c\cap F| = |M\cap F| - |M\cap E\cap F|$).
• And $M$ contains $20$ elements that are in $M$ but not in either $E$ nor $F$ (because $|M\cap E^c\cap F^c| = |M| - |M\cap E| - |M\cap F| + |M\cap E\cap F|$).

Similarly, $E$ contains $10$ elements that are also in $M$ and $F$; it contains $35$ that are in $M$ but not in $F$; $15$ that are also in $F$ but not in $M$; and therefore $15$ that in neither $M$ nor $F$.

And $F$ contains $10$ elements that are also in $M$ and $F$; $25$ that are in $M$ but not in $E$; and $15$ that are in $E$ but not in $M$; leaving $30$ that are in $F$ but not in $E$ nor $M$.

That is: \begin{align*} |M\cap E\cap F\;| &= 10;\\ |M\cap E\cap F^c\;| &= 35;\\ |M\cap E^c\cap F\;| &= 25;\\ |M\cap E^c\cap F^c\;| &= 20;\\ |M^c\cap E\cap F\;| &= 15;\\ |M^c\cap E\cap F^c\;| &= 15;\\ |M^c\cap E^c\cap F\;| &= 30;\\ |M^c\cap E^c\cap F^c\;| &= 0. \end{align*}