With respect taken to the very first problem posted by the OP (i.e. original poster), the approach taken by the original poster is sound, but needs refinement.
In my opinion, the informal winging it approach to detecting over-counting is error-prone. Safer is the formal approach of Inclusion-Exclusion. See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
I will follow the syntax in the 2nd link above.
Let $~S~$ denote the set of all possible permutations of the letters in AEINCB.
Let $~S_1~$ denote the subset of $~S~$ where the letters AE are fused together in some order.
Let $~S_2~$ denote the subset of $~S~$ where the letters AI are fused together in some order.
Let $~S_3~$ denote the subset of $~S~$ where the letters EI are fused together in some order.
Then, the desired computation is
$$\frac{ ~| ~S_1 \cup S_2 \cup S_3 ~| ~}{|~S ~|} ~: ~| ~S ~| = 6!.$$
Let $~T_1~$ denote $ ~| ~S_1 ~| + | ~S_2 ~| + | ~S_3 ~|.$
Let $~T_2~$ denote $ ~| ~S_1 \cap S_2 ~| + | ~S_2 \cap S_3 ~| + | ~S_1 \cap S_3 ~|.$
Let $~T_3~$ denote $ ~| ~S_1 \cap S_2 \cap S_3 ~|.$
Then,
$$| ~S_1 \cup S_2 \cup S_3 ~| = ( ~T_1 - T_2 ~) + T_3.$$
To compute $~S_1~$ note that there are $~2!~$ ways of internally permuting the letters AE into one fused unit. Then, since $~2~$ of the letters have been fused into one unit, there are $~5~$ units to permute, which can be done in $~5!~$ ways. Therefore,
$$| ~S_1 ~| = 2 \times 5!.$$
By considerations of symmetry,
$$| ~S_1 ~| = | ~S_2 ~| = | ~S_3 ~|.$$
Therefore,
$$T_1 = 3 \times 2! \times 5!.$$
To compute $~| ~S_1 \cap S_2 ~| ~$ you have to have both AE together and AI together. This requires that AEI all be fused together, with A being in the middle. So, you must fuse all $~3~$ vowels into one unit, with $~2!~$ internal permutations allowed, since the outer letters, E and I can appear in either order.
Then, with all $~3~$ vowels fused into $~1~$ unit, you have $~4~$ units to permute, which can be done in $~4!~$ ways.
Therefore,
$$| ~S_1 \cap ~S_2 ~| = 2! \times 4!.$$
By considerations of symmetry,
$$| ~S_1 \cap S_2 ~| = | ~S_1 \cap S_3 ~| = | ~S_2 \cap S_3 ~|.$$
Therefore,
$$T_2 = 3 \times 2! \times 4!.$$
To compute $~T_3~$ you are trying to enumerate all of the possible ways that AE are together, AI are together, and EI are together.
However, if AE are together and AI are together, then you must have A in between E and I, which makes it impossible for EI to be together.
Therefore,
$$T_3 = | ~S_1 \cap S_2 \cap S_3 ~| = 0.$$
So, the desired computation of the probability is
$$\frac{T_1 - T_2}{ | ~S ~| } = \frac{ ~( ~6 \times 5! ~) - ( ~6 \times 4! ~) ~}{6!}.$$
Dividing each portion of the above computation by $~4!~$ gives
$$\frac{ ~( ~6 \times 5 ~) - 6}{ 6 \times 5 } = \frac{4}{5}.$$
Two vowels together + 3 vowels together.
I can take any of 2 vowels out of 3 and arrange it in 3P2 ways multiplied by remaining letters which can be arranged in 5! ways (4 letters + one vowel group). Similarly followed same approach for 3 vowels together. My answer is 3P25! + 3!4! (3 remaining letters and 1 vowel group). But I feel there is an overlap between these two arrangements and not sure how to remove them. I have not been able to go beyond this for now ("AEINCBB" or "AEIINCB"). Thanks for promptly responding.
– Chandan Jul 12 '15 at 06:46