Given a word "aab", permutations are:
aab, aab, aba, aba, baa, baa
I need to get the number of permutations where characters don't repeat. So from the above permutations, I need to ignore those which has consecutive characters. So, I need to the result to be 2 which is (aba, aba).
How can I achieve this?
Another example: given word "aabb", I need to achieve 8 which is:
abab, abab, abab, abab, baba, baba, baba, baba
Let's try the example with four characters first, then you can try and see if you can mimic it for three characters.
You have "aabb" that you want to distribute into "_ _ _ _", two cases: one starting with $a$ another starting with $b$.
Case 1: Starting with $a$. This means that you have $2! \cdot 2! \cdot 1! \cdot 1! = 4$ way of permuting the characters.
Let's reason why this is the case. The first place can be one of two $a$'s available. So that's $2!$ ways of choosing it. Then the second place needs to be a $b$, which again you have two of, so $2!$ ways of putting a $b$ there. Then the third and fourth places, you've only got $1$ $a$ and $1$ $b$ left, so they both have $1!$ ways of being put there.
Case 2: Starting with $b$. This means you have $2!\cdot 2!\cdot 1!\cdot 1! = 4$ ways of permuting the characters.
In total, this means you have $4 +4 = 8$ ways of permuting them. Although, really, with experience, you're going to notice that splitting them into cases doesn't really matter, you could have simply exploited the symmetry present.
