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?:
Best Answer
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 $$\dfrac{4,032,000}{47,174,400} = \dfrac{10}{117}$$