Inclusion-Exclusion with 4 sets

Question

At MEXICANO tourism club, tourists can mix and match different activities while visiting Mexico. There are 49 tourists visiting Hel-Ha Park (H), 61 tourists visiting Pyramid of Tenochtitlan (T), 52 tourists visiting Chichen Itsa (C), and 28 visiting the National Museum of Anthropology (M). There are exactly 20 tourists for any two of these visits, 10 tourists for any three of them and 5 taking all of them. Using inclusion-exclusion, calculate how many tourists are doing at least one or more visits.

ANSWER = 105

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I don't understand why the answer to this question is 105. To me it should be 190 - 20 + 10 - 5 = 175

Answer 1

As there are $20$ tourists taking each pair of visits, the subtraction for two cases hould be $6 \cdot 20$ as there are ${4 \choose 2}=6$ pairs of visits that can be taken. Similarly the addition for three visits should be ${4 \choose 3} \cdot 10=40$ as the people who visited three things got counted three times in the single visit case and subtracted three times in the two visit case, so need to be added once. The count should be $190-6\cdot 20+4\cdot 10-5=105$

Answer 2

Each $3$ visits are made $10$ tourists, in which $5$ of them make all the $4$ visits. Similarly each $2$ visits are made by $20$ tourists in which $15$ make either all the visits or $3$ of them. So there are exactly $5$ tourists for any of $2$ visits ONLY (i.e. doesn't do $3$ visits).

I think now you can draw a venn diagram, those that visit ONLY (H) are $19$, ONLY (T) $31$, ONLY (C) $22$, ONLY (M) must be $8$ (if you adjust the number from $28$ to $38$)