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I just solved a TWIN problem for you under the link
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1126099.html
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1126099.html
This time the situation is very similar:
You are given 3 sets (the number of elements in three sets M, H and G);
You are given the numbers of elements in their in-pair intersections (M n H), (M n G) and (H n G);
and finally, you are given the numbers of elements in the triple intersection (M n H n G).
They ask you about the number of elements in the union of the three sets (M U H U G).
The same formula (*) works as in my previous post:
n(M U H U G) = n(M) + n(H) + n(G) - n(M n H) - n(M n G) - n(H n G) + n(M n H n G). (*)
you only need to substitute the given data and calculate:
n(M U H U G) = 50 + 60 + 70 - 25 - 28 - 32 + 21 = 116.
Answer. The number of peoples who failed at least one subject is 116.
Solved.
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This time I will not prove the formula (*) here: it is just proved under the referred link.
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For many other solved similar problems see the lessons
- Counting elements in sub-sets of a given finite set
- Advanced problems on counting elements in sub-sets of a given finite set
- Challenging problems on counting elements in subsets of a given finite set
in this site.