In how many ways can we select $6$ cards from a pack of $52$ cards such that we get all $4$ suits? Number of ways to select $1$ card from each suit $= 13^4=28561$
Total cards remaining $=52-4=48$
Cards remaining in each suit $=13-1=12$
Now I have to choose $2$ more cards and this can happen in two ways. First, either both cards are selected from the same suit or both cards are selected from two different suits.
Number of ways to choose $2$ cards from the same suit $=^{12}C_2+^{12}C_2+^{12}C_2+^{12}C_2 = 66 \times 4 = 264$ 
Number of ways to choose $2$ cards from two different suit = $=(^{12}C_1 \times ^{12}C_1\times^{12}C_0\times^{12}C_0 ) \times ^{4}C_2 = 864$
So the total number of ways should be $=28561 \times (264 + 864) = 32,216,808$
But it seems that this answer is wrong. Please help me to understand what I am doing wrong.
Thanks in advance !!!
 A: JMoravitz has provided you with two nice solutions to the problem.  To understand where you made your mistakes, we will consider how your method compares with solving by cases.
Solution:
There are two cases:

*

*three cards are drawn from one suit and one card apiece is drawn from each of the other three suits

*two cards each are drawn from two of the suits and and one card apiece is drawn from each of the other two suits

Three cards are drawn from one suit and one card apiece is drawn from each of the other three suits:  There are four ways to select the suit from which three cards are drawn and $\binom{13}{3}$ ways to draw three cards from that suit.  For each of the other three suits, there are $13$ ways to select a card from that suit.  Hence, the number of such cases is
$$\binom{4}{1}\binom{13}{3}\binom{13}{1}^3$$
Two cards each are drawn from two of the suits and and one card apiece is drawn from each of the other two suits:  There are $\binom{4}{2}$ ways to select the suits from which two cards are drawn.  For each of those two suits, there are $\binom{13}{2}$ ways to select two cards from that suit.  For each of the other two suits, there are $13$ ways to select a card from that suit.  Hence, the number of such cases is
$$\binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2$$
Total:  Since these cases are mutually exclusive and exhaustive, the number of six-card hands which contain at least one card from each suit is
$$\binom{4}{1}\binom{13}{3}\binom{13}{1}^3 + \binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2 = 8,682,544$$
What errors did you make?
As lulu indicated in the comments, you counted hands multiple times.
By designating a particular card as a representative of each suit and then selecting additional cards, you counted each case in which three cards are drawn from one suit and one card each is drawn from each of the other three suits three times, once for each way you could have designated one of the three cards from which three cards are drawn as the representative of that suit.
Let's illustrate this with the following example:  Suppose the hand you selected is $5\clubsuit, 8\clubsuit, J\clubsuit, \color{red}{5\diamondsuit}, \color{red}{7\heartsuit}, K\spadesuit$.  Your method counts this hand in three ways.
\begin{array}{c c c c c}
\text{clubs} & \text{diamonds} & \text{hearts} & \text{spades} & \text{additional cards}\\ \hline
5\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 8\clubsuit, J\clubsuit\\
8\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 5\clubsuit, J\clubsuit\\
J\clubsuit & \color{red}{5\diamondsuit} & \color{red}{7\heartsuit} & K\spadesuit & 5\clubsuit, 8\clubsuit\\
\end{array}
By designating a particular card as a representative of each suit and then selecting additional cards, you count each hand in which two cards each are drawn from two suits and one card each is drawn from the other two suits four times, two times each for each way you could designate one of the two cards from the suits from which two cards are drawn as the card of that suit.
Let's illustrate this with the following example:  Suppose the hand you selected is $4\clubsuit, \color{red}{5\diamondsuit}, \color{red}{9\diamondsuit}, \color{red}{Q\heartsuit}, 8\spadesuit, 10\spadesuit$.  Your method counts this hand in four ways:
\begin{array}{c c c c c}
\text{clubs} & \text{diamonds} & \text{hearts} & \text{spades} & \text{additional cards}\\ \hline
4\clubsuit & \color{red}{5\diamondsuit} & \color{red}{Q\heartsuit} & 8\spadesuit & \color{red}{9\diamondsuit}, 10\spadesuit\\
4\clubsuit & \color{red}{5\diamondsuit} & \color{red}{Q\heartsuit} & 10\spadesuit & \color{red}{9\diamondsuit}, 8\spadesuit\\
4\clubsuit & \color{red}{9\diamondsuit} & \color{red}{Q\heartsuit} & 8\spadesuit & \color{red}{5\diamondsuit}, 10\spadesuit\\
4\clubsuit & \color{red}{9\diamondsuit} & \color{red}{Q\heartsuit} & 10\spadesuit & \color{red}{5\diamondsuit}, 8\spadesuit\\
\end{array}
Note that
$$\color{red}{3}\binom{4}{1}\binom{13}{3}\binom{13}{1}^3 + \color{red}{4}\binom{4}{2}\binom{13}{2}^2\binom{13}{1}^2 = \color{red}{32,216,808}$$
A: You can do this directly by breaking into cases:

*

*$3+1+1+1$ case:  Pick which suit has three cards picked from it.  Pick what those three cards are.  Then for each remaining suit in canonical order, pick what card appears from that suit.  $\binom{4}{1}\binom{13}{3}\cdot 13\cdot 13\cdot 13$


*$2+2+1+1$ case:  Pick which two suits have two cards picked for them.  For those suits in canonical order, pick what cards those were that appeared for the suits.  For the remaining two suits in canonical order, pick what card appears from that suit.  $\binom{4}{2}\binom{13}{2}\binom{13}{2}\cdot 13\cdot 13$
Adding these cases together gives a final total of $8682544$
This is... fine... for such a small problem.  But if we were to have gotten ten cards?  Or 15 cards?  You do not want to be breaking into cases since there will be many such cases to consider and it will be easy to make an arithmetic mistake or miss a case or make a typo etc...
Another perhaps more elegant approach, one which does not suffer the same problems, is that of using the Inclusion-Exclusion Principle.
We consider the question of how many hands exist which are missing at least one of the suits and we subtract that from our total.  Letting $A$ be the set of hands missing $\clubsuit$, and similarly defining $B,C,D$ as missing $\diamondsuit,\heartsuit,\spadesuit$ respectively... our original question is asking us to find $|A^c\cap B^c\cap C^c\cap D^c| = |\Omega|-|A|-|B|-|C|-|D|+|A\cap B|+\dots - |A\cap B\cap C|-\dots+|A\cap B\cap C\cap D|$
Thanks to all of the symmetry involved here, this simplifies quite a bit.  For our six-card hand this would be:
$$\binom{52}{6}-4\binom{39}{6}+\binom{4}{2}\binom{26}{6}-4\binom{13}{6}+0 = 8682544$$
This is, of course, the same answer as before.  This also can easily be modified to answer the question of if we had a $k$ card hand... just all of the bottom numbers in the binomial coefficients which were $6$ you now have $k$'s instead.  No messy casework required.
