# Combinatorics Inclusion-Exclusion

At one school, three computer languages, Basic, FORTRAN, and Pascal are taught. Suppose that for each language 27% of the students know that language, for each pair of languages 12% of the students know that pair of languages, and 7% of the students know all three languages. How many of the students do not know any computer language?

I am not sure how to start this but I think I use inclusion-exclusion method?

Yup, inclusion/exclusion is at work here. Drawing a Venn diagram will prove particularly fruitful. Be mindful of intersections and what is in what: you should get something like this:

• Since the intersection of two circles is $$12\%$$, we can conclude, since the intersection of all three is $$7\%$$, that the blue regions (the pairwise intersections excluding the intersection of all) is $$12\% - 5\% = 7\%$$

• Similarly, since each circle has $$27\%$$ total, we subtract the blue/white regions of each circle to find that the red regions are $$10\%$$.

• What we seek is the gray: those not accounted for by the circles. Sum up all the percentages from each region and subtract from $$100\%$$ to get the desired answer.

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

Using this formula, number of students not knowing any language can be obtained. Hope it helped you.

• This is close but there are 4 sets in the problem not only 3. See the comment I made up under the question. – NoChance Apr 3 at 10:00