We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?
Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 \choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.
I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)
The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.
The cats can now be arranged in $3!$ ways.
So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$
Where have I gone wrong?
from Hot Weekly Questions - Mathematics Stack Exchange
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