Combinatorics Name Generating So I'm supposed to figure out how many names an alphabet consisting of A,L,I,S and T can generate. Each name must consist of $2$ vowels and $5$ consonants, begin and end with a consonant, two vowels can't be adjacent and two adjacent consonants must be different. I calculated that there are $2,592$ names possible by looking at the $6$ different forms the names can take and add up the number of names each form generates. However, I was wondering if this problem can be done using the Rule of Product. Since there are $3$ forms with $2$ isolated consonants and $3$ forms with $1$ isolated consonant, I don't believe there is. 
 A: Observations: All vowels will have a consonant to the right of it.  The first letter will always be a consonant.  (Notice: this forces the condition that the last letter is a consonant as well as forces the condition that vowels are seperate)
Let us construct the word in the following way:
Let the first letter be a consonant.  Arrange $V_C,V_C, C,C$.
With the first letter as a constant, we have arrangements such as $CV_CCCV_C$ corresponding to a word where the first, third, fourth, fifth, and seventh letters are consonants.
"Two adjacent consonants must be different" can be satisfied instead by the phrase "Every consonant is different than the letter to its left"
From left to right, pick a letter to be used for the available spot.  Although the order of the steps might change depending on which arrangement it is, there will still be a total of three times where you pick a consonant with no restriction, two times where you pick a consonant such that it cannot be the same as the consonant to its left, and two times where you pick a vowel.
Each time you pick a consonant with no restriction, you will have three choices.
Each time you pick a consonant with restriction (has a consonant to its left), you will have only two choices.
Each time you pick a vowel, you will have two choices.
The total is then $\underbrace{\binom{4}{2}}_{\text{arrangements of}~C,V_C}\cdot \underbrace{3\cdot 3\cdot 3}_{\text{consonants with no consonant to left}}\cdot \underbrace{2\cdot 2}_{\text{consonants with consonant to left}}\cdot \underbrace{2\cdot 2}_{\text{vowels}} = 3^4\cdot 2^5=2592$
