Simple probability question with two colored balls? If I have $m$ black balls and $n$ white balls. Whats the probability that an arrangement of them will have no 2 black balls placed next to each other?
I am thinking for if $n \le m-1$ then its always 1/total arrangements.
If say we have 2 black and 1 white ball... theres only 1 way to do this, bwb.
For 3 black and 2 white, theres also only one way to do this... bwbwb.. so 1/10?
So answer should just be $\frac{n!m!}{(n+m)!}$?
What if $n > m$?
 A: Think of the balls as individuals, each with her own student number.  If the balls are placed in an urn, and drawn out one after the other, to be placed in a row, then the assumption of labelling does not affect the probability, it just makes the calculation clearer.
There are $(m+n)!$ equally likely ways to arrange the balls in a row.
Now let us count the number of arrangements in which no $2$ black balls are next to each other. Write down a series of $n$ $X$'s, to indicate the ultimate positions of the white balls, like this:
$$X\qquad X\qquad  X\qquad X\qquad  X\qquad X\qquad  X\qquad X\qquad $$
These determine $n+1$ "gaps" (counting the two at the ends). We must choose $m$ of them to slip a black ball into. 
This can be done in $\dbinom{n+1}{m}$ ways. For each such way the $m$ black can be arranged in the chosen positions in $m!$ ways, and then the white in $n!$ ways, for a total of
$$\binom{n+1}{m}m!n!.$$
For the probability, divide by $(m+n)!$.
Note that the answer is technically correct even for $m \gt n+1$, if we use the convention that $\binom{a}{b}=0$ if $a\lt b$. 
A: Note: I am assuming, as in my experience is usually intended in such problems, that the $m$ black balls are mutually indistinguishable, and so are the $n$ white balls. If the balls are individually identifiable, the calculations must be modified to take order into account, and it’s easier to insert black balls into the string of white balls, as in André’s answer.
There are $\binom{m+n}m$ distinguishable arrangements of the balls in a row. If $r(m,n)$ is the number that do not have two adjacent black balls, the desired probability is $$\frac{r(m,n)}{\binom{m+n}m}\;,$$ so the heart of the problem is computing $r(m,n)$.
Imagine laying out the $m$ black balls in a line, with spaces between adjacent balls: with $m=7$, for instance, you get the arrangement $_B_B_B_B_B_B_B_$. There are $m-1$ internal slots for white balls and $2$ slots on the ends. We must put at least one white ball into each of the internal slots, but we needn’t put any into either end slot. Put one white ball into each of the internal slots:
$$_BW_BW_BW_BW_BW_BW_B_$$
You now have $n-(m-1)=n-m+1$ white balls to distribute arbitrarily amongst the $m+1$ available slots. Each possible distribution gives you exactly one of the $r(m,n)$ arrangements with no two consecutive black balls, so $r(m,n)$ is just the number of ways of distributing $n-(m-1)=n-m+1$ white balls amongst $m+1$ slots. This is a standard stars-and-bars problem; you’ve probably already seen such problems, but if not, the link gives both a formula and a pretty clear explanation of where it comes from.
