Answer: Check out the diagram below for the filled in boxes
14 goes in the first box (inside A, but outside B)
7 goes in the overlapping circle regions
5 goes in the third box (inside B, outside A)
3 goes in the box outside of the circles
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Explanation:
[tex]n(A \cup B) = 26[/tex] means there are 26 items that are in A, B or both.
n(A) = 21 means there are 21 items in A
n(B) = 12 means there are 12 items in B
We don't know the value of [tex]n(A \cap B)[/tex] which is the number of items in both A and B at the same time. This is the intersecting or overlapping regions of the two circles. Let [tex]x = n(A \cap B)[/tex]
It turns out that adding n(A) to n(B), then subtracting off the stuff they have in common, leads to n(A u B) as shown below.
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[tex]n(A \cup B) = n(A) + n(B) - n(A \cap B)\\\\26 = 21+12 - x\\\\26 = 33 - x\\\\x+26 = 33\\\\x = 33-26\\\\x = 7\\\\n(A \cap B) = 7\\\\[/tex]
So there are 7 items in both regions.
This means there are [tex]n(A) - n(A \cap B) = 21 - 7 = 14[/tex] items that are in set A only. In other words, 14 items are in circle A, but not in circle B.
Notice how the values 14 and 7 add back up to 14+7 = 21, which represents everything in set A.
Similarly, there are [tex]n(B) - n(A \cap B) = 12 - 7 = 5[/tex] items that are in circle B, but not in circle A. The values 5 and 7 in circle B add to 5+7 = 12, matching with n(B) = 12.
The notation n(A') means the number of items that are not in set A. We're given n(A') = 8. We already know that 5 is outside circle A. So if 5+y = 8, then y = 3 must be the missing value for the box that is outside both circles.
Again the diagram is posted below with the filled in values.
