The goal is to fill out all of the empty spaces with corresponding values.
We're given the following
n(A intersection B intersection C)=
so we'll write "28" in the region that corresponds to where A, B and C overlap simultaneously. That's the very middle region
Let's update Figure 1 to get Figure 2, which now has "28" in the proper place
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Now let's focus on the intersections of two sets only. Let's start with n(A intersection B)=
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Let's repeat those steps for n(A intersection C)=35 now. We know that 35 would go in the region between A and C, but we have to kick out the 28 already in the very center: 35-28 = 7. So 7 goes in the overlap between A and C but it's outside of set B. Figure 4 reflects this update
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The last pairing of sets is n(B intersection C)=37 meaning that 37-28 = 9 items are in set B and C but not in set A. Figure 5 shows the update
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Now to the individual sets. We'll focus on each set one at a time. Set A has 79 items as n(A)=79 states. But we've already accounted for a lot of members in this set already. Figure 5 shows the values 7, 28 and 12 in the circle A. Add them up: 7+28+12 = 47. So there are 47 members in circle A so far. There are 79 total meaning there are 79-47 = 32 items that are in set A but not in set B nor in set C. Figure 6 shows the update
Note as a check you can add up all the values in circle A (that figure 6 displays). Doing so gets you: 32+7+28+12 = 79
which confirms we have the proper value in the right place.
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Repeat the steps done above, but now for set B. We're told that n(B)=75 but we see from figure 6 the values in circle B are 12, 28 and 9 which add to 12+28+9 = 49. So this means n(B) - 49 = 75 - 49 = 26 goes in the region that is inside B but not in the other circles as Figure 7 shows
Check: Refer to figure 7. Add up the values in circle B: 12+28+9+26 = 75 which confirms we have the right value.
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Now onto set C. We have n(c)=59. Add up the values in circle C so far: 7+28+9 = 44. Then subtract this result from the n(C) value: n(C)-44 = 59-44 = 15. So 15 goes in the circle C but it is not inside circle A nor is it in circle B
Check: Refer to figure 8. Add up the values in circle C: 7+28+9+15 = 59. So we have the right value
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We're almost done. We have one more region to deal with. This is the region that is not in any of the circles but it is still inside the rectangle U. This represents the items that are not in any of the sets A,B,C. Add up all the values you see in Figure 8: 32+12+26+7+28+9+15 = 129
This meeans there are 129 items that are in set A, set B, or set C or in some combo/overlap of the sets. We're told that n(U) = 135. There are 135 items in the universal set. So there must be 135-129 = 6 members that are outside the three circles. Write '6' anywhere you wish as long as it's outside all three circles. I've done so in the lower left corner as shown in Figure 9
Figure 9 shows the final result of what you should have. The numbers can slide around as long as they stay in the proper region.