combinatronics solving METHOD doubt regarding combination A word consists of 9 letters of 4 consonants and 5 vowels. Three letters are chosen at random. Find the probability that more than one vowels is chosen. 
At first I thought this was the answer:
$$1- (\frac{4C3}{9C3} + \frac{4C2*5C1*2}{9C3})$$
i multiplied the second part by 2 since I assumed that the vowels could be chosen first or second and the order might change.
After seeing the answer I realised that the answer should be:
$$1-(\frac{4C3}{9C3} + \frac{4C2*5C1}{9C3})$$
I got how to solve this question after looking at the answer. But I could not quite understand why my previous method was wrong. Some questions require me to multiply the combinations by a number depending on how many ways something can be done in, and some don't. What is the underlying concept?
 A: The probability that no vowels are chosen is $\binom 43$.
Let's say exactly one vowel is picked, and two consonants are picked. This happens in $\binom 51 \times \binom 42$ ways, because we can choose the vowel out of $5$ given, and $2$ consonants out of the four given. 
The order in which the consonants are picked does not matter, because we are taking combinations here, and not permutations : we do not care about the order in which the consonants are picked. 
But why is that? That, is because the statement of the question asks : "the probability of one or more vowels occuring in our choice", and it starts with "three letters are chosen". There is no mention of the word order : it does not say  something like "the second one should be a vowel and the third should be a consonant" or something like that. That's why, all our choices are made as combinations i.e. we don't worry about order.
So the crucial point is choice  : when you are asked to choose, you don't take order into account. It would not have mattered whether you chose the vowels first, second or third, it is a property of the collection of letters that is being asked for, so a choice is being made, therefore omitting the multiplication by $2$.
