Checking whether to use a permutation or combination So I'm studying for my Discrete Math final tomorrow and wanted to make sure I had a clear understanding of permutations versus combinations - it's my understanding that you use permutations when order matters - and combinations doesn't care about order.
That being said, I've come across a problem and wanted to make sure my answers were appropriate, and clear up a step.
"You have a group of 10 individuals, six women and four men. If you have four positions on a council (chairperson, co-chair, treasurer, secretary) how many possible combinations can you have if:


*

*The chairperson is a woman

*The Chairperson and co-chair are women?
These are my steps for both questions:
{Number 1}
$P(6,1) \rightarrow  \frac{6!}{5!}$
$ P(9,3) \rightarrow \frac{9!}{6!}$
$ = \frac{6!}{5!} \cdot \frac{9!}{6!}$
{Number 2}
$P(6,2) \rightarrow  \frac{6!}{4!}$
$ P(8,3) \rightarrow \frac{8!}{5!}$
$ = \frac{6!}{4!} \cdot \frac{8!}{5!}$
Now I haven't done the final calculation but I'm not 100% sure my numbers are right, since it seems like the actual answers I would get would be very large - could someone braek down whether my process is correct here? Am I right in using a permutation versus combination formula? Am I supposed to multiply the two results or add them?
 A: You are entirely right in saying that 

you use permutations when order matters - and combinations doesn't
  care about order.

but keep in mind that the number of permutations is simply the number of combinations multiplied by the number of ways you can order the objects, or in symbols, $P(n, k) = C(n, k) \cdot k!$. To arrange some objects, you first choose some of them, then move them around.
A good thing to remember in combinatorics is that, almost always, when you use the word "and" to describe a situation, you have to multiply, and when you use "or", you have to add. In your problem, note that the conditions (after paraphrasing) read:


*

*The chairperson is a woman and the other three posts can be chosen anyhow.

*The chairperson is a woman and the co-chair is a woman and the other two posts can be chosen anyhow


You have solved the second condition in a simpler way, doing $P(6, 2)$ directly instead of $P(6, 1) \cdot P(5, 1)$ which is the same thing.
If the condition was (parentheses denote precedence of 'and' and 'or'): "(The chairperson is a woman and the other posts can be chosen anyhow) or (The chairperson is a woman and the co-chair is a woman and the other posts can be chosen anyhow)", then you would have to do
$$P(6, 1) \cdot P(9, 3) + P(6, 1) \cdot P(5, 1) \cdot P(8, 2)$$
Hope this clarifies things!
