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Bullet
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Bunuel
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Bullet
Twenty-four dogs are in a kennel. Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair. There is only one dog that is black with a short tail and long hair. Two of the dogs are black with short tails and do not have long hair. Two of the dogs have short tails and long hair but are not black. If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?

I'll write solution later.

The answer is indeed 3.

12+6+15-4-x-2(1) = 24
which implies x = 3.
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Attachment:
24-Dogs in Kannel.GIF
24-Dogs in Kannel.GIF [ 10.16 KiB | Viewed 27973 times ]

9 - x + 2 + 1 + 1 + 2 + x + 12 - x = 24

27 - x = 24

x = 3
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I am curious to know why the answer is not 7.

Total = Group1 + Group2 + Group3 - Group1/2 - Group1/3 - Group2/3 + 2(Group1/Group2/Group3)

24 = 12+6+15-2-2-x+2
24 = 33-4-x+2
24 = 29-x+2
22 = 29-x

x=7

Can someone point out my mistake?
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I am curious to know why the answer is not 7.

Total = Group1 + Group2 + Group3 - Group1/2 - Group1/3 - Group2/3 + 2(Group1/Group2/Group3)

24 = 12+6+15-2-2-x+2
24 = 33-4-x+2
24 = 29-x+2
22 = 29-x

x=7

Can someone point out my mistake?

ManU,

I reckon the formula that you are using is not correct -

The formula to use is -
Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

If you put the values in the above, the answer comes out to be 3.
See my earlier post above.
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This wasnt complex but very time consuming. Good problem, though.
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I am completely baffled by this question.

When I solve it using the formula:
# of items total = (a + b + c) - (ab + ac + bc) + (abc)

I get 6

When I solve it the same way as Bullet, I get 3.

Can someone please suggest, where the mistake is?

Thanks.
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When do we use these formula?>
total = A+B+C - (AnB+BnC+AnC) + Neither + (AnBnC)
total = A+B+C - (sum of exactly two groups) - 2(AnBnC) +Neither
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Bunuel,
This is great. Thank you soooo much :-)
I hope not to make mistakes again with overlapping sets problem.
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Quote:
As we are given members of exactly 2 groups then use second formula:

Thank you, Bunuel. I knew how to derive the formula. But I've only now realized that we have the number of exactly two groups and not two groups including the intersection of three sets. Very careless of me.

Thank you again.
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