Sorry, the answer is 13. I know that this question can be found in the forum, but I'd like to know if there's a way (such as using the standard overlapping set equation) to attack the problem other than by drawing out each of the three groups!
Thanks so much!
You can use the formula for 3 sets:
n(AUBUC) = n(A) + n(B) + n (C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
n(AUBUC) = 8 + 8 + 8 - 5 - 5 -5 + 4 = 13
Though, why would you not try using the Venn diagram for 3 overlapping sets. This question is relatively straight forward so the formula is no problem, but with the Venn diagram, it doesn't matter how much you twist the question, the answer is still apparent....
I am attaching it in case you would like to give it a look...
Ques2.jpg [ 10.5 KiB | Viewed 1295 times ]
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