The three sets $A, B,$ and $C$ each have $2018$ elements. The intersection of any two of the sets has $201$ elements. The intersection of all three sets has $20$ elements. How many elements are there in the union of the three sets?
If it is saying there are 20 elements in the intersection of the three sets, isn't it saying there are 20 numbers that are not repeated throughout the sets? So wouldn't the union be the same? Is there another way to solve?
On
|A|=|B|=|C| = 2018 |A∩B|=|B∩C|=|C∩A|= 201 |A∩B∩C|= 20 Using Inclusion-ExclusionPrinciple, we have |A∪B∪C|=|A|+|B|+|C| -|A∩B|-|B∩C|-|C∩A|+|A∩B∩C| |A∪B∪C|= 3.2018-3.201+20=5471 You can draw Venn diagram for more clarity. Or you can think logically as follows: Union of n number of sets kills the elements which are shared among more than one set. These 20 elements are shared by these three sets, so we will kill the repetition of them and count them only once. Similarly, 201 elements are shared by two sets and 20 out of them are shared by three therefore, we need to kill those 201 elements keeping in mind that 20 out of them had already been killed. That gives us a number of 181. Now, each set has 2018 elements but it shares 181 with two different sets and 20 as the common in all. Hence, elements that each set has amd only they have it are 2018-181-181-20= 1636. Hence, the total elements in union is 3.1636+3.181+20=5471
Hint:
Apply the Inclusion-Exclusion Principle:
$ |A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$.