probability of who will be selected "In an office there are 3 secretaries, 4 accountants, and 2 receptionists. If a committee of 3 is to be formed, find the probability that one of each will be selected?
Attempted Solution:
First attempt: (3/9)(4/9)(2/9) = 8/243
Second attempt: (3/9)(4/8)(2/7) = 1/21
Don't if either of these are correct or not. Any help would be greatly appreciated to point me in the right direction. Thank you. 
 A: The standard way to do this is to say there are $\binom{9}{3}$ equally likely ways to choose $3$ people from the $9$. And there are $\binom{3}{1}\binom{4}{1}\binom{2}{1}$ ways to choose one of each kind. For the secretary can be chosen in $\binom{3}{1}$ ways. For each such choice, the accompanying accountant can be chosen in $\binom{4}{1}$ ways. And once this has been done, the receptionist can be chosen in $\binom{2}{1}$ ways.
For the probability, divide "favourables" by "total." We get
$$\frac{\binom{3}{1}\binom{4}{1}\binom{2}{1}}{\binom{9}{3}}.$$
Another way: Imagine choosing the people one at a time. We find the probability of choosing a secretary, then an accountant, then a receptionist.
The probability the first person chosen was an S is $\frac{3}{9}$. Given this happened, the probability the next chosen person was an A is $\frac{4}{8}$. and given this happened, the probability the next person chosen was an S is $\frac{2}{7}$, for a probability of $\frac{3}{9}\cdot\frac{4}{8}\cdot\frac{2}{7}$.
But there are several other ways we could end up with one of each, for example A then S then R. If you calculate, this ends up having the probability $\frac{4}{9}\cdot\frac{3}{8}\cdot\frac{2}{7}$. This is the same number as the one previously obtained. 
There are in total $3!=6$ ways we can end up with one of each. so the required probability is 
$$6\cdot \frac{3}{9}\cdot\frac{4}{8}\cdot\frac{2}{7}.$$
The first way is (after you get used to it) easier. 
