The probability of bricks arranged randomly. If someone have $4$ red bricks and $8$ blue bricks and arranges them randomly in a circle,
what is the probability that two red bricks are not side by side?
Is the sample space  $\frac{12!}{4!\,8!} \,=\, 495$ ?
and the answer $P(\text{Reds non-adjacent}) \:=\:\frac{70}{495} \:=\:\frac{14}{99}$ ?
 A: Suppose there $r$ red bricks, and $b$ blue bricks. Counting clock-wise, let 
$$
   (k_1, k_2, \ldots, k_r)
$$ 
be the tuples, representing the number of blue bricks between two red bricks. The number of favorable configurations is:
$$
    N_F = \sum_{k_1=1}^{b-r+1} \sum_{k_2=1}^{b-r+1} \cdots \sum_{k_r=1}^{b-r+1} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{t-t^{b-r+2}}{1-t} \right)^r = [t^{b-r}] \left( \frac{1-t^{b-r+1}}{1-t} \right)^r = [t^{b-r}] \left(1-t\right)^{-r} = \binom{b-1}{r-1}
$$
and the number of total configurations:
$$
  N_T = \sum_{k_1=0}^{b} \sum_{k_2=0}^{b} \cdots \sum_{k_r=0}^{b} \delta_{k_1+k_2+\cdots+k_r, b} = [t^b] \left( \frac{1-t^{b+1}}{1-t} \right)^r= [t^b] \left(1-t\right)^{-r} = \binom{b+r-1}{b}
$$
Hence the probability:
$$
   p =\frac{N_F}{N_T} = \frac{\binom{b-1}{r-1}}{\binom{b+r-1}{b}} = \frac{b! (b-1)!}{(b-r)! (b+r-1)!}
$$
With $b=8$ and $r=4$, we have $N_F = \binom{7}{3} = 35$ and $N_T = \binom{11}{8} = 165$, thus
$$
   p = \frac{N_F}{N_T} = \frac{35}{165} = \frac{7}{33}
$$
A: There are ${12\choose 4}=495$ ways to choose the 4 places to put the red bricks.
From my answer here, the number of ways to choose 4 non-consecutive places is 
${12\over 8}{8\choose4}=105.$ (Put $d=4$ and $n=12$). The desired probability is then 
$105/495=7/33=.2121$.
