Combinatorics exam question: number of possible actions when order matters I'm preparing for my exam later this week, and I've come across a question of which I do not understand the answer.
The question reads thus:
A stock market dealer trades only with shares "A". Six times per trading day he performs one of the following three actions:


*

*he buys a number of shares "A", or 

*he sells a number of shares "A", or 

*he does not buy nor sell.
Because the actions have effect on the share price "A", the sequence of the actions are of interest. How many possibilities are there to have exactly four selling actions on a trading day (and thus two other actions)?
I thought the answer would be in the form of $\frac{n!}{(n-k)!}$ or in this case $\frac{6!}{(6-4)!}$ but this appears to be off by a factor 2. I suspect this has something to do with the fact that there are 'two other actions', thus multiplying the total amount of outcomes, though I would not be able to explain it fully. Could someone tell me how to make sense of this?
 A: Order matters here, so $$\text{sell, sell, sell, sell, buy, buy}$$ is a different string (set of actions) than say $$\text{buy, buy, sell, sell, sell, sell}$$ So combinations are not the right tool since they ignore the order of the events. That said, lets use (indeed) combinations to solve it. We split the process first in two steps, count the possible ways to perform each step and then multiply to obtain the result, by the Multiplication Principle (or Product Rule). So,


*

*Choose $4$ actions out of the $6$ to be $\text{sell}$. You can do this in $\dbinom{6}{4}$ ways. So, here we specified which actions will be "sell". For example $$\text{___, ___, sell, sell, sell, sell}$$ 

*Choose $2$ actions to fill in the remaining two actions. You have two choices $(\text{buy, nothing})$ for each remaining action, giving $2\times2=4$ to choose the remaining actions. Here we took into account that order matters, and so we distinguish $(\text{buy, nothing})$ and $(\text{nothing, buy})$ (the other $4$ being "sell") as two different scenarios.


Applying the multiplication principle, you get that there are $$\dbinom{6}{4}\times 4=60$$ ways to perform $4$ "sell's" and $2$ other actions in the day.  
