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Here is the question.

Each of the 90 students participated in at least one of the three track events A, B and C. If 20 students participated in A, 40 students participated in B, and 60 students participated in C, and if 5 students participated in all three events, how many students participated in at least two of these events?

Now, I applied the simple inclusion exclusion principle here -

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| \\ |A \cap B| + |A \cap C| + |B \cap C| = |A| + |B| + |C| + |A \cap B \cap C| - |A \cup B \cup C|$

This gives 35 as the result. However the current result is 20 (as per attached explanation). What am I doing wrong here?

enter image description here

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The Venn diagram is there to help. Formulas tend to look all alike: it is good to look at the picture and see what is counted by $|A\cap B|+|A\cap C|+|B\cap C|$. –  André Nicolas Sep 1 '12 at 0:10
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1 Answer 1

up vote 1 down vote accepted

The problem is that $|A \cap B| + |A \cap C| + |B \cap C|$ isn’t the number who participated in at least two events: every student who participated in all three events got counted in $|A\cap B|$, again in $|A\cap C|$, and yet a third time in $|B\cap C|$. Those five students have therefore been counted three times each. In the notation of the attached explanation, $$|A \cap B| + |A \cap C| + |B \cap C|=(b+e)+(d+e)+(f+e)=b+d+f+3e\;.$$

If you want the number who participated in at least two events, you want to count each of those five people once, not three times, so the correct answer is $35-2\cdot5=25$. If you want the number who participated in exactly two events, you don’t want to count those five people at all, so the correct answer is $35-3\cdot5=20$. The attached explanation gives the answer to this second question, not to the one that you asked at the beginning of the post.

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Cool. Thanks a lot! –  user957 Aug 31 '12 at 23:20
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