If you have $9$ books, $4$ from Shakespeare, $2$ by Dickens, and $3$ from Tolstoy. How many ways can you arrange the books such that no two Tolstoy books are next to each other?
I would like to use Stars and Bars (not that any other method is bad but I'm trying to learn this method in particular.)
This is my method. If you wish to bash me, do so at once. Since no Tolstoy can touch, there are $7$ spaces ($6$ bars) that a Tolstoy can go into. The total arrangements is then $9!$ (with or without stars and bars). At this point I am lost.
What would be the proceeding steps and considerations?
The way you propose, using the $6$ "non-Tolstoy" books as bars, is the simplest, but for variety, I am giving a solution making the Tolstoy books as bars.
Firstly we consider as if books by the same author are identical.
With $T$ representing Tolstoy books, there are four gaps, in which the remaining $6$ books must be placed, with at least one book in each of the middle two gaps.
$- T - T - T- $
Pre-place one book in each of the middle two gaps,
now we need to place the remaining $4$ books any way we like in the $4$ gaps.
By stars and bars, number of ways = $\binom{4+4-1}{4-1} = \binom73$
And, of course, if each book is distinct, the answer becomes $\binom736!3!$