Find the number of ways to arrange so that at least two are followed 
Suppose there is a sentence containing only sequences of three characters and nothing more. The three characters are $X,Y,Z$ and it is given that $X$ has occurred $a$ times, $Y$ has occurred $b$ times and $Z$ has occurred $c$ times in the sentence. What is the probability that a $X$ will  be followed by a $Y$ at least $2$ times$?$

This problem looks a bit complicated to me, so I decided to break it into some parts. At least $2$ times means, all cases$-$$($exactly $0$ time$+$exactly $1$ time$)$
The number of all cases are simply $$\binom{a+b+c}{a}\binom{b+c}{b}\binom{c}{c}$$
For the exactly $0$ times I'm able to think of a logic but it is hard to explain. I think that we should first select a $Z$ and fix it. Then we arrange  $b$ $Y's$ and $(c-1)$ $Z's$ on one side of the fixed $Z$ and on other side put all the $X$. One more is to put all the $X's$ between two $Z's$ and then do the rest of the arrangement. So you see I'm not been able to think of all such cases. One more is put all $Y's$ and then arrange rest such that no $X$ goes behind any $Y$.
For exactly $1$ I have similar incomplete cases. How to find all possible cases in each sub problem$?$
Any help is greatly appreciated.

As this answer has now been accepted, I should mention that it’s unnecessarily complicated and the stars-and-bars argument provided by Daniel Mathias in a comment below is much more elegant.

You can proceed as in this nice answer by A.J. to What is the probability of 2 named cards appearing sequentially in a randomly shuffled deck if suits are ignored?.
First, to find the probability that no $X$ is followed by a $Y$, arrange the $X$s and $Z$s in some way and then insert the $Y$s. For the first insertion, there are $a+c+1$ equiprobable slots where the $Y$ can go, and $a$ of them are behind an $X$. If we ever put a $Y$ behind an $X$, it’s over (since we can’t prevent that $X$ from being followed by a $Y$ by inserting further $Y$s behind it). The probability to survive the first insertion is $\frac{c+1}{a+c+1}$, to survive the second insertion $\frac{c+2}{a+c+2}$, and so on. Thus, the probability that no $X$ is followed by a $Y$ is
$$
\frac{c+1}{a+c+1}\cdot\frac{c+2}{a+c+2}\cdots\frac{c+b}{a+c+b}=\frac{(c+b)!(a+c)!}{c!(a+c+b)!}\;.
$$
For the probability that exactly one $X$ is followed by a $Y$, we need to replace one of the factors by its complement, and the numerator of all factors after that is increased by $1$ because we can now put $Y$s behind the $X$ that we put a $Y$ behind. Thus, we get an additional factor $c+b+1$ in the numerator, and if the $k$-th $Y$ is inserted after an $X$ the factors $c+k$ and $c+k+1$ are replaced by $a$. Thus the probability that exactly one $X$ is followed by a $Y$ is
$$
\frac{(c+b+1)!(a+c)!}{c!(a+c+b)!}\cdot a\left(\frac1{(c+1)(c+2)}+\frac1{(c+2)(c+3)}+\cdots+\frac1{(c+b)(c+b+1)}\right)\;.
$$
The sum in parentheses telescopes because
$$\frac1{(c+k)(c+k+1)}=\frac1{c+k}-\frac1{c+k+1}\;,$$
so the sum is $\frac1{c+1}-\frac1{c+b+1}=\frac b{(c+1)(c+b+1)}$, for a probability
$$
ab\cdot\frac{(c+b)!(a+c)!}{(c+1)!(a+c+b)!}\;.
$$
Subtracting these two from $1$ yields the probability that at least two $X$s are followed by a $Y$ as
$$
1-\frac{(c+b)!(a+c)!}{c!(a+c+b)!}\left(1+\frac{ab}{c+1}\right)\;.
$$
