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Total combinations = 3!*3!
6*6 = 36
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Bunuel
In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the second, fourth, and sixth cages?

A. 12
B. 18
C. 36
D. 72
E. 720

Our arrangement must be as follows:

D - C - D - C - D - C

For the first cage, we have 3 possibilities because there are 3 available dogs for the first cage. For the second cage, we have 3 possibilities for the 3 available cats. For the third cage, we have 2 available dogs (because one dog was already placed in the first cage). Similarly, for the fourth cage, we have 2 available cats. Finally, for the fifth cage, we have the one remaining dog, and for the sixth cage, the one remaining cat. .

So, the number of ways to arrange the animals is 3 x 3 x 2 x 2 x 1 x 1 = 36.

Answer: C
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