We first place the non-math books. There are $4$ choices for the first book, $3$ choices for the second book, $2$ choices for the third book, and $1$ choice for the last book. Then we have to put the two math books between the four non-math books such that there is at least one non-math book between the two math books. We see there is a total of $5$ openings created by the four non-math books. So the first math book has $5$ choices, and the second math book has $4$ choices.

So the total number of ways the books can be placed is $4\times3\times2\times1\times5\times 4 =\boxed{480}.$
Question:
Suppose that I have $6$ different books, $2$ of which are math books. In how many ways can I stack my $6$ books on a shelf if I do not want the math books to be next to each other?
Answer:
Let's think step by step.

user
480
Suppose that I have $6$ different books, $2$ of which are math books. In how many ways can I stack my $6$ books on a shelf if I do not want the math books to be next to each other?
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