Here are the clues:
Clues
A. 8 would not go to the park.
B. 7 would not go to the beach.
C. 11 would not go to the family cottages.
D. 3 would go to neither a park or beach.
E. 4 would go to neither a beach nor the family cottage.
F. 6 would go to neither a park nor the family cottage.
G. 2 would not go to the park or a beach or a family cottage.
H. 1 would go to all three places.
Let's re-word all the clues and consider the members as
non-park goers, non-beach goers, and non-cottage goers:
Clues
A. There are 8 non-park goers.
B. There are 7 non-beach goers.
C. There are 11 non-cottages goers.
D. There are 3 non-park and non-beach goers.
E. There are 4 non-beach and non-cottage goers.
F. There ate 6 non-park and non-cottage goers.
G. There are 2 non-park, non-beach and non-cottage goers.
H. There is 1 who is neither a non-park, non-beach, or non-cottage goer.
So lets make a three-set Venn Diagram with three mutually overlapping
circles.
All the non-park goers are in the NP circle.
All the non-beach goers are within in the NB circle.
All the non-cottage goers are within in the NC circle.
Each little letter s through z represents the
number of members in the category or categories
NP, NB, and/or NC the region that letter is in:
s = the number of NP only
t = the number of NP and NB, but not NC
u = the number of NB only
v = the number of NP and NC but not NB
w = the number of NP,NB, and NC
x = the number of NB,NC, but not NP
y = the number of NC only
z = the number of not NP, not NB, and not NC
So now we can look at the clues and determine:
Clues
A. There are 8 non-park goers. So s+t+v+w=8
B. There are 7 non-beach goers. So t+u+w+x=7
C. There are 11 non-cottages goers. So v+w+x+y=11
D. There are 3 non-park and non-beach goers. So t+w=3
E. There are 4 non-beach and non-cottage goers. So w+x=4
F. There ate 6 non-park and non-cottage goers. So v+w=6
G. There are 2 non-park, non-beach and non-cottage goers. So w=2
H. There is 1 who is neither a non-park, non-beach, or non-cottage goer. So z=1
From G we have w=2
From H we have z=1
From D we have t+w=3 and since w=2, t+2=3 and so t=1
From E we have w+x=4 and since w=2, 2+x=4 and so x=2
From F we have v+w=6 and since w=2, v+2=6 and so v=4
From A we have s+t+u+v=8, so s+1+4+2=8, so s=1
From B we have t+u+w+x=7, so 1+u+2+2=7, so u=2
From C we have v+w+x+y=11, so 4+2+2+y=11, so y=3
So we have a total of s+t+u+v+w+x+y+z = 1+1+2+4+2+2+3+1 = 16
Answer: 16 Montesano family members.
Edwin