Counting Question Been wrestling with the following counting question for about an hour. I will explain my reasoning for counting in the question and I request any BETTER WAY to make this calculation or corrections if my counting is wrong.
"If each coded item in a catalog begins with 3 distinct letters followed by 4 distinct   nonzero digits, find the probability of randomly selecting one of these coded items with the first letter a vowel and the last digit even."
I first calculate the sample space:
26 * 25 * 24 * 9 * 8 * 7 * 6
= 47,174,400

This is done using the simple "placeholder" technique for counting.
Now I must calculate the special case such that the first letter is a vowel and the last digit even. The "placeholder" technique confuses me here.
5 * 25 * 24 * 9 * 8 * 7 * 4 = 6,048,000

So probability  is 6048000/47,174,400.
I can't  help but think this is wrong. What about the case where the code has a vowel BEFORE the last vowel, for example. Does this not make the calculation the following:
 5 * 25 * 24 * 9 * 8 * 7 * 3

Because one vowel has already been seleccted BEFORE the last vowel... there are fewer choices.
Should I need to add up all the permutations in which there is a vowel previous to the last vowel? That seems agonizing... what theorems and rules can I get both the correct and easiest path to the answer?
After some Google-ry, I found this explanation for the solution in a textbook. How do they get this? What witchery is this? Why is it so different than the other answers?:

 A: Your calculation for letters is just fine. I think you confused yourself with the rest of your question. Recall, we have $3$ distinct letters followed by $4$ distinct numbers. So vowels need not come into play at the end of the sequence. If you meant "even numbers": we can choose to fill the even number first (even though it appears last), leaving $8$ remaining numbers distinct from that the even number chosen to fill the second "digit" slot, then $7$ numbers to fill the third "digit" slot, then $6$ for the last digit. Overall,  the combinations with first letter a vowel, last number an even number is $$5\cdot 25\cdot 24 \cdot 8\cdot 7\cdot 6\cdot 4 = 4,032,000$$ 
So the probability of selecting one of the desired combinations from among all the combinations in the sample space is given by $$\dfrac{4,032,000}{47,174,400} = \dfrac{10}{117}$$
A: Let $\mathcal E$ be the event that the first character is a vowel and $\mathcal F$ be the event that the last digit is even. 
$\begin{align*}
P[\mathcal E \cap \mathcal F] &= P[\mathcal E]P[\mathcal F | \mathcal E]\\
&= P[\mathcal E]P[\mathcal F] &\text{(the events are clearly independent)}\\
&= \frac{5}{26} \times \frac 4 9\\
&= \frac {10}{117}
\end{align*}$
A: For your first question, you did fine with the letters.  You have five ways to select the first vowel, then 25 and 24 for the next two letters because you don't care whether they are vowels or consonants.  For the numbers, the problem is you don't know how many evens were chosen in the first three.  It is better to select the last digit first, as you know four evens are available, so the count of numbers is $4*8*7*6$
When you say a vowel before the last vowel, is that just two vowels?  You have four choices for the layout of vowels and consonants:  VVC, VCV, CVV, VVV.  Assuming they are all acceptable, the first three all have the same number of combinations, so we can calculate the first and multiply by 3:  $3*5*4*21$, where the $5,4$ are the number of vowels available and the $21$ is the number of consonants.  VVV is then $5*4*3$, so the total number of letter orders is $3*5*4*21+5*4*3=1320$.  If we just did $3*5*4*24$ (figuring that we had $24$ letters available after we picked the two vowels) we would count VVV three times, once for making each letter the "wild card".
