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. 

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. 
