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Re: Three dogs and three cats are to be arranged in a circle so that no tw [#permalink]
ThatDudeKnows wrote:
BrentGMATPrepNow wrote:
Three dogs and three cats are to be arranged in a circle so that no two dogs are next to each other. If two arrangements are considered different only when the positions of the animals are different relative to each other, how many arrangements are possible?

(A) 12
(B) 18
(C) 24
(D) 36
(E) 60


If no two dogs are next to each other, we always need a cat between any two dogs. So the only way to seat these animals is DCDCDC.

On circle questions, it's easiest to start by simply choosing a spot for one of the elements.
Let's put a cat at seat 1.
Seat 2 must be a dog. How many options do we have for seat 2? There are three dogs and we can choose any of them, so three options.
Seat 3 must be a cat. How many options do we have for seat 3? We already have one cat seated, so two options.
Seat 4 must be a dog. How many options do we have for seat 4? We already have one dog seated, so two options.
Seat 5 must be a cat. How many options do we have for seat 5? We already have two cats seated, so one options.
Seat 6 must be a dog. How many options do we have for seat 6? We already have two dogs seated, so one options.
3*2*2*1*1 = 12

Answer choice A


Can you please explain why you didn’t count 3 options for the first cat?

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Re: Three dogs and three cats are to be arranged in a circle so that no tw [#permalink]
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The answer will be the number of ways in which one group can be arranged in a circle \((n-1)!\), multiplied by the number of ways in which the other group can be arranged \(n!\)

Total arrangements of cats in a circle: \((3-1)! = 2\)

Total arrangements of dogs: \(3! = 6\)

Three dogs and three arranged in a circle so that no two dogs are next to each other: \(2*6 = 12\)

Answer A
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Re: Three dogs and three cats are to be arranged in a circle so that no tw [#permalink]
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Re: Three dogs and three cats are to be arranged in a circle so that no tw [#permalink]
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