3 different math books, 2 different physics books are arranged on a bookshelf. In how many ways can they be arranged if no 2 math books are together?

Question

The book gave the answer as $12$.

This is how I approached it:

We need the arrangements in the form

math - physics - math - physics - math

or else two math books will be adjacent. So, we choose $2$ positions for the two physics books to be placed and multiply it by $2!$, since either position can have either book. Next, we see that there are three remaining spaces for the math books, so we multiple by $3!$ giving us:

$$ 2 \times 2 \times 3 \times 2 \times 1 $$

However, since we chose the positions once for the physics books, and then another for the math books, we must divide by $2$ so we don't double count. Does this seem correct?

Answer 1

I think you overcomplicated it a little bit.

Your argument that the arrangement should be

$$ MPMPM $$

is correct, $M$ standing for math books and $P$ standing for physics.

Now you could basically look at the number of permutations between the $3$ math and $2$ physics books. Which gives

$$ 3!\cdot 2!=12 $$

different arrangements.

You can think of this last step as "we know that the books will be in this order $MPMPM$" and this is not changed by swapping two math books for example.

We can in a way decompose this into two parts

$$ M\ \ M\ \ M $$ and $$ P\ \ P $$

and look at the number of ways the individual parts can be arranged and then using product rule to reassamble them.

P.S: I don´t really understand your last paragraph about overcounting. Would you elaborate on it a bit?

Answer 2

There won't be able double counting in this case but you seem to have made a simple error, you have put $2$ for each physics book but once one is in place, there is no choice for the other. Starting from the left, your choices at each stage are $3, 2, 2, 1, 1$ and hence the answer if $12$.