Condon wrote:
Could someone explain to me why it is necessary to double the amount of parents in all groups?
Thanks
This post tells you how you get the formula for n(Exactly two sets) and others:
http://www.veritasprep.com/blog/2015/10 ... ping-sets/This is our standard formula:
Total = 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(No Set)39 = 21 + 18 + 19 - (n(A and B) + n(B and C) + n(C and A)) + 5 + 0
(n(A and B) + n(B and C) + n(C and A)) = 24
Now let’s see how we can calculate the number of people in exactly two sets.
n(A and B) includes people who are in both A and B and it also includes people who are in A, B and C. Because of this, we should remove n(A and B and C) from n(A and B) to get n(A and B only). Similarly, you get n(B and C only) and n(C and A only), so adding all these three will give us number of people in exactly 2 sets.
n(Exactly two sets) = n(A and B) – n(A and B and C) + n(B and C) – n(A and B and C) + n(C and A) – n(A and B and C).
Therefore:
n(Exactly two sets) = n(A and B) + n(B and C) + n(C and A) – 3*n(A and B and C)n(Exactly two sets) = 24 - 3*5 = 9
Answer (D)
In other words, you can substitute from formula 2 into formula 1 to get
n(A and B) + n(B and C) + n(C and A) = n(Exactly two sets) + 3*n(A and B and C)
Put in formula 1:
Total = n(A) + n(B) + n(C) – (n(Exactly two sets) + 3*n(A and B and C)) + n(A and B and C) + n(No Set)
Total = n(A) + n(B) + n(C) – n(Exactly two sets) - 2*n(A and B and C)) + n(No Set)
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