5 black, 7 red, 9 blue, and 6 white marbles. I'm having trouble finding how many ways there is to arrange 5 black, 7 red, 9 blue, and 6 white marbles to find the probability that every white marble is adjacent to at least one other white marble.
If you could help me out that would be great!
 A: There are, in general, $n!$ ways to arrange $n$ objects. So you'd have $(5+7+9+6)!$ but then as all the black marbles are identical, their permutations shouldn't be counted. For every 'good' permutation, you also have another $5!7!9!6!$ that only differ from it by permutations of same colored marbles. Thus the result is $\frac{(5+7+9+6)!}{5!7!9!6!}$.
For the probability, consider that whites must be in pairs or triples. This is equivalent to solving the previous problem for $3$ and $2$ white marbles. Notice that the two cases overlap when all $6$ marbles are in one group.
A: First find the total probability space of the marbles. I assume the marbles of each color are otherwise indistinguishable and so order doesn't matter.
$N_{all} = \frac{(5+7+9+6)!}{5!\cdot7!\cdot9!\cdot6!}$
Second, create an expression for the number of combinations of the black, red, and blue marbles, this will be the number of contexts for the white marbles .
$N_{contexts} = \frac{(5+7+9)!}{5!\cdot7!\cdot9!}$
Our total probability is going to equal...
$P = \frac{N_{contexts} \cdot N_{whites}}{N_{all}} = \frac{21!6!}{27!} \cdot N_{whites}$
Next think about how many ways every white marble can be adjacent to at least one other white marble: (2,2,2), (2,4), (3,3), and (6,0)
The (6,0) case is the easiest because there is only one moveable entity. There are (5+7+9)+1 possible locations within the other marbles (5+7+9 locations before another marble and an extra 1 location at the very end).
$N_{1} = 22$
The (3,3) case and the (4,2) case have equal combinations because they both have two moveable entities. The first set of white marbles can be in (5+7+9)+1 possible locations, and that leaves (5+7+9) locations for the second set. Order shouldn't matter, so we get...
$N_{2} = \frac{(5+7+9+1)(5+7+9)}{2} = 231$
For the (2,2,2) case (our last one!), we have (5+7+9)+1 locations for our first pair, (5+7+9) for our second, and (5+7+9)-1 for our third. Again order shouldn't matter, so we get...
$N_{3} = \frac{22!}{19!\cdot3!} = 1540$
Now, we say that $N_{whites} = N_1 + 2 \cdot N_2 + N_3 = 2024$ and plug that value into the original formula and we get...
$P = \frac{21!6!}{27!} \cdot 2024 = .00683...$
So the final probability is about 0.7%.
A: We describe one way to count the number of ways to arrange the balls. 
Let the number of ways to place the white marbles be $W$. Each of these ways  leaves $21$ empty slots. Let $N$ be the number of ways to fill these slots. Then our answer is $WN$.
Finding $N$ is probably a standard problem for you. 
Finding $W$ is more complicated. We can split into cases. First count the number of ways to place the white marbles so they are all together.  
Then count the $4$-$2$ splits (and the $2$-$4$ splits). Then count the $3$-$3$ splits, and finally the $2$-$2$-$2$ splits. 
The counting of, for example, the $4$-$2$ splits is not hard. Imagine the non-whites lined up. Then there are $22$ gaps between them, including the end "gaps." Choose $2$ of these.
For the probability, divide by the number of ways to place the balls with no restriction. 
A: A simple way is to compute the complement.
The 21 "non-white" marbles have 22 "interstices" (including ends), so the white marbles can be placed in C(22,6) "non-adjacent" ways and the others permuted in 21!/(5!7!9!) ways in their positions
Pr = 1 - [C(22,6)*21!/(5!7!9!)]/[27!/(5!7!9!6!)]
= 1 - C(22,6)*6!21!/27!
