Find the number of elements in A1A2A3A1A2A3 if there are 9898 elements in A1A1,

Question

Find the number of elements in A1A2A3A1A2A3 if there are 9898 elements in A1A1, 10041004 elements in A2A2and 1008810088 elements in A3A3 in each of the following situations:
(a) The sets are pairwise disjoint.

(b) A1A2A3A1A2A3.

(c) There are 2020 elements common to each pair of sets and 55 elements in all three sets

(1 point) Find the number of elements in A UA2 UA3 f there are 98 elements in A1, 1004 elements in A2 and 10088 elements in A3 in each of the following situations: (a) The sets are pairwise disjoint. (b) A1 C A2 C A3 (c) There are 20 elements common to each pair of sets and 5 elements in all three sets.

Explanation / Answer

a) If the sets are pairwise disjoint, then they share no elements. As such, the union of the sets has as many as the sum of the elements in each set. So, 10088 + 98 + 1004 = 11190 elements.

b) If A1 is a subset of A2, then the union of the two is in fact just A2 itself (since every element in A1 is also in A2, combining the two of them leaves it as just A2. To illustrate this point consider, how many days are there if you combine the group of all of the days in february 2005 and all of the days in 2005? 365). And indeed if A2 is a subset of A3, then the union of the two is just A3. So it has as many elements as there are in A3, specifically 10088 elements
c) We can see that there are 20 shared between each pair, 55 among all three, and each circle totals to the right amount. You can add them up either by taking 10088 - 60 that were counted once extra - 55 that was counted twice extra = 10088 - 60 - 55 = 9973 elements

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