When to use Permutations and Combinations So I am very confused about when to use combinations and permutations in questions that ask me to find the amount of ways something can be ordered. 
For example, I was trying to answer a question that asked to find get $5$ kinds of cards (aces, twos, threes, etc) in a $5$ card hand. I thought that this would be a permutation $P(13,5)C(4,1)^5$, since I had read that the ways to get a full house ($2$ of one type and $3$ of another) was $P(13,2)C(4,3)C(4,2)$, and thought it would work the same way for this problem, but it ended up being a combination of $C(13,5)C(4,1)^5$.
Why is this the case? And how can I identify when to use combination or permutation?
Thanks,
Kyle
 A: If the order of the objects or the cards matters you need to use permutation.
If the order of the objects doesn't matter, you need to use combination.
In your example, any of the five cards can be picked randomly, where the order does not matter, so you use combination. 
Hope this helps.
A: The other two answers correctly mentioned the difference between order mattering and not mattering for permutations and combinations respectively. I however, wanted to shed some light on the implied problems of the OP which were:

*

*How to find the total number of $5$-of-a-kinds in a poker hand

*How to find the total number of full houses in a poker hand

For the first, as was previous mentioned, the order of the hand of five distinctly valued cards does not change the hand we have i.e. a hand of $A,2,3,4,5$ all fo hearts is identical to $5,4,3,2,A$ all of hearts. As such, to choose $5$ distinctly valued cards need employ $C(13,5)$. Now however, we have to realize that each card can come in four different suit, and thus for each of the five cards we chose with $C(13,5)$ we have $4$ options to choose from. This leads immediately to the thought of $C(4,1)$ for five cards yiedling the given answer of $C(13,5)*(C(4,1))^5$.
Now for the second. I actually quite dislike the answer given in the OP, that of $P(13,2)C(4,2)C(4,3)$ as I am of the impression that it is hiding the real permutation in the $P(13,2)$ term. We know that a full house consists of $2$ values of card, $2$ of the one and $3$ of a second. So exaclty like in the $5$-of-a-kind example we can choose $2$ of the $13$ possible values in a standard deck of $52$ cards to determine the number of ways in which choosing these two values is possible (NOTE that some may disagree that clearly there are $13$ options for the first value in a full house and $12$ for the other and thus $P(13,2)$ is correct. And I agree that it gives the correct value I just argue that it is arrived at via misleading reasoning based on the order mattering definitions differentiating $P$ and $C$). I would argue that the order of these two chosen values is not relevant and therefore calls for a $C$ to be calculated. The note above's arguement would have you say that not only are you choosing two of the $13$ values for your full house but you are simultaneously choosing which will be your group of $3$ and which your group of $2$, and though this is technically correct I believe it skips a step and is the reason the OP was confused. $C(13,2)$ is the clearer way because, like the $5$-of-a-kind example we are merely choosing cards. We care not that we chose $(A,K)$ versus $(K,A)$ After this choice, we can ask now how many ways can I pick my grouping of $3$ and my grouping of $2$? Well one would have $2$ options, i.e. the $2$ values just chosen in $C(13,2)$, to choose the grouping of $3$ and $1$ option for the grouping of $2$. Because the choice of these two determines which value will require $3$ cards and which $2$ in our full house we are effectively distinguishing between a full house of $(A,A,K,K,K)$ and $(A,A,A,K,K)$ with this permutation of $P(2,1)$. Then finally we can zoom in individually on our pair and $3$-of-a-kind and choose our suits $3$ of them for our $3$-of-a-kind choice and $2$ for our pair yielding $C(4,3)*C(4,2)$. This culminates in a solution of $C(13,2)P(2,1)C(4,3)C(4,2)$ which of course is the same value as the OP's just arrived at in a, in my opinion, clearer way as it stays a bit truer to the definitions that are the answer to the over-arching question of what distinguishes a $P$ from a $C$. Hope this clears it up!
A: Just to throw another 2 scenarios that might make sense when to use combination or permutation: 


*

*You are the Coach of a 12 basketball players, you need to choose 5 to play the game, does it matter in what order you select them? No! so that combination.

*You're running a 100 meters sprint with another 4 guys on the lane. Does it matter who get first, second etc? Yes! so that permutation 
