Let W, X, Y, Z be subsets of {1, 2, . . . , 100} such that W ∩X = ∅, W ∩Y = ∅ and X ∩Y = ∅. Use the inclusion-exclusion principle to write down an expression for |W ∪ X ∪ Y ∪ Z|

Based on the question, I'm puzzled on what Z is as it is not mentioned in the question. How can you write an expression for |W ∪ X ∪ Y ∪ Z| when Z is not given?

  • $\begingroup$ The question you have given only asks for an expression for $\lvert W \cup X \cup Y \cup Z \rvert$, not to actually compute the cardinality. $\endgroup$ – Mattos May 7 at 11:41
  • $\begingroup$ If you have an expression for $W \cup X \cup Y$ can you find an expression for $W \cup X \cup Y \cup Z$? $\endgroup$ – Jay May 7 at 11:44
  • $\begingroup$ Just work out $|W\cup X\cup Y\cup Z|$ by means of inclusion/exclusion and the data in your question. $\endgroup$ – drhab May 7 at 11:45
  • $\begingroup$ It is good to add a Venn diagram. $\endgroup$ – NoChance May 7 at 12:09

$Z$ is mentioned in the question; it is one of the subsets. It doesn’t matter that we have no extra information about it, though, because we are just asked to write an expression for $|W \cup X \cup Y \cup Z|$ rather than actually compute its numerical value.

The inclusion-exclusion principle states, in this case, that

$$|W \cup X \cup Y \cup Z| = |W| + |X| + |Y| + |Z| - |W \cap X| - |W \cap Y| - |W \cap Z| - |X \cap Y| - |X \cap Z| - |Y \cap Z| + |W \cap X \cap Y| + |W \cap Y \cap Z| + |X \cap Y \cap Z| + |W \cap Y \cap Z| - |W \cap X \cap Y \cap Z| $$

In this case we know that $|W \cap X| = |W \cap Y| = |X \cap Y| = 0$ and so we in fact have

$$|W \cup X \cup Y \cup Z| = |W| + |X| + |Y| + |Z| - |W \cap Z| - |X \cap Z| - |Y \cap Z| + |W \cap X \cap Y| + |W \cap Y \cap Z| + |X \cap Y \cap Z| + |W \cap Y \cap Z| - |W \cap X \cap Y \cap Z| $$

We can remove any of the intersections that contain these intersections to give

$$|W \cup X \cup Y \cup Z| = |W| + |X| + |Y| + |Z| - |W \cap Z| - |X \cap Z| - |Y \cap Z| $$

If there is any more data in your question, you can use then this to make this expression more precise.

  • $\begingroup$ Is it correct to say that if $W \cap Y = 0 \implies W\cap Y \cap Z=0$? $\endgroup$ – NoChance May 7 at 11:55
  • $\begingroup$ Yes. you are correct. Thank you. (I think you either mean the cardinalities are zero or that zero should be $\emptyset$ though) $\endgroup$ – 雨が好きな人 May 7 at 12:00

Let $T=W\cup X\cup Y$. Then \begin{align*} |W\cup X\cup Y\cup Z|&=|T\cup Z|\\ &=|T|+|Z|-|T\cap Z|\\ &=|W\cup X\cup Y|+|Z|-|(W\cap Z)\cup(X\cap Z)\cup(Y\cap Z)|. \end{align*} The inclusions-exclusion principle is used in going from the first line to the second line. Now, we are given $W,X$, and $Y$ are mutually disjoint. This also implies that $W\cap Z$, $X\cap Z$, and $Y\cap Z$ are mutually disjoint. And so: \begin{align*} |W\cup X\cup Y|&=|W|+|X|+|Y|\\ |(W\cap Z)\cup(X\cap Z)\cup(Y\cap Z)|&=|W\cap Z|+|X\cap Z|+|Y\cap Z|. \end{align*} Thus, $$ |W\cup X\cup Y\cup Z|=|W|+|X|+|Y|+|Z|-|W\cap Z|-|X\cap Z|-|Y\cap Z|. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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