Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have three sets A, B and C satisfying the following conditions:

  • $ \#(A\cap B) = 11$
  • $ \#(A\cap C) = 12$
  • $ \#(A\cap B\cap C) = 5$ What is the minimun cardinality of A?

What I did was this:

(11 - 5) + (12 - 5) = 13.

But I'm not sure if I have to substract the 5 twice or only once. And the fact that I don't know if B and C are disjoint gets me confused =/

share|cite|improve this question
Why did you calculate $(11-5)+(12-5)$? – Chris Eagle Dec 30 '12 at 20:17
Also, you do know if $B$ and $C$ are disjoint. – Chris Eagle Dec 30 '12 at 20:18
You should really draw this... – Karolis Juodelė Dec 30 '12 at 20:21
I did this: $ ( \#(A\cap B)-\#(A\cap B\cap C)) - ( \#(A\cap C)-\#(A\cap B\cap C))$ – Victor Jose Arana Rodriguez Dec 30 '12 at 20:23
You must subtract the $5$ only once. – P.. Dec 30 '12 at 20:27
up vote 5 down vote accepted

\begin{align} \#(A\cap (B\cup C)) &= \#((A\cap B)\cup (A\cap C))\\ &=\#(A\cap B)+\#(A\cap C)-\#(A\cap B\cap C)\\ &=11+12-5=18. \end{align} Note that $A$ cannot be fewer than $18$ since $A\cap(B\cup C)\subseteq A$, thus $\#(A)\geq18$.

Now, if we can find an example, where $\#(A)=18$, we'll have minimized the cardinality of $A$. For such an example, let $A=\{1,\ldots,18\}$, $B=\{1,\ldots,11\}$, and $C=\{7,\ldots,18\}$.

share|cite|improve this answer

Note that $$A = (A\cap B^c \cap C^c) \bigcup (A \cap B \cap C^c) \bigcup (A\cap B^c \cap C) \bigcup (A \cap B \cap C)$$ In the above, we have written $A$ as the union of four disjoint sets. Hence, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B \cap C^c) + \#(A\cap B^c \cap C) + \#(A \cap B \cap C) \,\,\,\,\,\,\,\, (\star)$$ Now note that we can write $A \cap B$ and $A \cap C$ as a disjoint union as follows. $$(A \cap B) = (A \cap B \cap C^c) \bigcup (A \cap B \cap C)$$ and $$(A \cap C) = (A \cap B \cap C) \bigcup (A \cap B^c \cap C)$$ Hence, we get that $$\#(A \cap B) = \#(A \cap B \cap C^c) + \#(A \cap B \cap C)$$ and $$\#(A \cap C) = \#(A \cap B \cap C) + \#(A \cap B^c \cap C)$$ Rearranging, we get that $$\#(A \cap B \cap C^c) = \#(A \cap B) - \#(A \cap B \cap C)$$ $$\#(A \cap B^c \cap C) = \#(A \cap C) - \#(A \cap B \cap C)$$ Plugging the above two in $(\star)$, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B) + \#(A \cap C) - \#(A \cap B \cap C) \,\,\,\,\,\,\, (\dagger)$$ Plugging in the given values in $(\dagger)$, we get $$\#A = \#(A\cap B^c \cap C^c) + 11 + 12 - 5 = \#(A\cap B^c \cap C^c) + 18$$ Note that $\#X \geq 0$ for any set $X$ and hence, the minimum value of $\#A$ is when $\#(A\cap B^c \cap C^c) = 0$, which gives us $$\#A = 18$$ Note that the minimum is attained when $A \subseteq B \cup C$.

share|cite|improve this answer
@Clayton $A = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\}$, $B = Z^{-} \cup \{0,1,2,3,4,5,6,7,8,9,10,11\}$ and $C = \{7,8,9,10,11,12,13,14,15,16,17,18,19,20,\ldots\}$ – user17762 Dec 30 '12 at 21:00
@Clayton There is no $6$ in $C$. It is enough to start from $7$. Number of numbers between $a$ and $b$ including both is $b-a+1$. – user17762 Dec 30 '12 at 21:12
@Clayton $A \cap B \cap C = \{7,8,9,10,11\}$. – user17762 Dec 30 '12 at 21:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.