Probability of drawing five distinct cards without a pair I am trying to determine the probability of obtaining five distinct cards(in terms of type, so only 1 two card, 1 three card, 1 four card, 1 five, 1 six for example).
I believe that I have found the answer by taking $(1) * 12/13 * 11/13 * 10/13 * 9/13 = .4165...$.
What would be the standard way of solving this problem?(assuming this is correct, which it may not be)
$note:$ I am not sure if Combinatorics is the correct tag, but it came up with (subtag?) 'counting'. Also note, this is homework, so if possible, please help me better understand the structuring for answering these questions(so I can get the rest right).

Your answer is correct and the standard approach if there are a very large number of cards of each type in the deck, so the removal of the first card doesn't change the chance to get to get a second one of that type.  It would also be correct if the cards are put back into the deck after each draw.  More usually, the cards drawn are not available for subsequent draws.  It is important to specify which way the sampling is done.  Sampling without replacement in a standard deck of $52$ cards, the chance the second matches the first is only $\frac 3{51}$, not $\frac 1{13}$, because there are only three matching cards left out of the remaining $51$.  The corresponding calculation to yours is then $1\cdot \frac {48}{51}\cdot \frac {44}{50}\cdot \frac {40}{49}\cdot \frac{36}{48}=\frac {2112}{4165}\approx 0.5071$
