Probability that when 5 cards are drawn from a standard deck that cards are drawn from only 3 of the suits without having three cards from one suit Compute the probability that when drawn 5 cards at random from a randomly shuffled deck of 52 cards you are drawn cards from only 3 of the suits without having three cards from one suit, in other words 2 cards from one suit, 2 from another, and 1 from a third. For this problem I know how to get started but because the probability is reliant on two variables, I'm not exactly sure how to calculate the probability in this instance so any help would be of great help. I currently have little to no progress on this so if you know where I can start or how to get started or if you can just help with the whole thing, anything and everything is helpful as of now.
Apparently I need to explain why I'm asking this question so the question was a question have thinking about for a while and never could get any traction on it.
 A: For $5$ cards belonging to $3$ different suits, following are the possibilities:

*

*$2$ cards belonging to $1$ suit, $2$ cards belonging to $2$nd suit and $1$ card belong to $3$rd suit.

*$3$ cards belonging to $1$ suit, $1$ cards belonging to $2$nd suit and $1$ card belong to $3$rd suit.

The question rules out the 2nd possibility. So, for 1. we calculate the probability as follows:
$\begin{align}P(E) &= \frac{(^4C_2\ ^{13}C_2 \ ^{13}C_2)\ (^2C_1\ ^{13}C_1)}{^{52}C_5} \\ &= \frac{\frac{4\cdot3}2\cdot\frac{13\cdot12}2\cdot\frac{13\cdot12}2\times2\cdot13}{\frac{52\cdot51\cdot50\cdot49\cdot48}{5\cdot4\cdot3\cdot2\cdot1}} \\ &= \frac{1521}{4165} \\ &= 0.3652\end{align}$

Here, $(^4C_2\ ^{13}C_2 \ ^{13}C_2)$ represents that we first choose any 2 suits of the 4 suits and then draw $2$ cards each from the same suit. 
And $(^2C_1\ ^{13}C_1)$ represents that we now choose a suit from the remaining $2$ suits and then draw a card from it.
Finally, $^{52}C_5$ represents all the possibilities if choosing $5$ cards from the deck of $52$ cards.
