An Unlucky Situation

Suppose we have a set $A=\{1,...,n\}$ . Among all possible the subsets of $A$ , $S_m(A)=\{s_1, s_2, ... , s_m\}$ , such that $|S_m(A)|=m$ , we need to find the ones in which none of the elements are consecutive, i.e. $|s_i-s_j |\geq 2$ for all $i \neq j$ . From a classic stars and bars argument it can be proved that the number of such subsets is

$\displaystyle |\{S_m(A) \ : \ |s_i-s_j |\geq 2, \ \forall i \neq j\}|=\binom{n-m+1}{m}$

for $\displaystyle m \in \bigg[0,\biggl \lfloor\frac{n+1}{2}\bigg \rfloor\bigg]$ .

Hence, out of $n$ throws, the probability that no two consecutive consonants show up is

$\displaystyle \mathbb{P}(\text{no two consecutive consonants})=\sum_{m=0}^{\lfloor\frac{n+1}{2}\rfloor} \binom{n-m+1}{m} 4^m 2^{n-m}$

For $n=13$ , we get

$\displaystyle \mathbb{P}(\text{no two consecutive consonants})=\sum_{m=0}^{7} \binom{13-m+1}{m} 4^m 2^{13-m}=\frac{3641}{531441}=\frac{11 \cdot 331}{3^{12}}=\frac{a\cdot b}{c^d}$

And

$\displaystyle a \cdot b \cdot c \cdot d =\boxed{131076}$

Let's imagine 13 empty blocks where we will put either Vowel(V) or Consonant(C). So there can be many Combinations of the relative position of V and C.

As we have 4 options of C and 2 options of V :

Case 1: 0 Consonant and 13 vowels: 1x2^13

Case 2: 1 Consonant and 12 vowels: 13C1 x 2^12 x 4

Case 3: 2 Consonant and 11 vowels: As there can be no consecutive consonants, we have to put those 2 consonants in any two of the 12 gaps out or in between the 11 vowels.So the permutation will be: 12C2 x 2^11 x 4^2

Case 4: 3 Consonant and 10 vowels: 11C3 x 2^10 x 4^3 . . . . .

Case 8: 7 consonants and 6 vowels: 7C7 x 2^6 x 4^7

Adding them up we get: 2^13 x 10923 So the Probability is :(2^13 x 10923)/(6^13) = 10923/3^13 = (11 x 331)/(3^12)

so a= 11, b= 331, c=3, d=12

and so, a x b x c x d= 131076