There are two ways of enumerating the number of possible $(6)$ card hands, from a $(52)$ card deck, when the cards are drawn without replacement, and where you want to reject any $(6)$ card hands that involve all 4 suits.
One way is to enumerate the number of $(6)$ card hands that do involve all 4 suits, and deduct that from
$$\binom{52}{6} = \frac{(52)!}{(6!)[(52-6)!]}.$$
The alternative method is the direct approach, which is detailed below. I have chosen this approach because, while it is more difficult, it is also more educational.
You have a $(3)$ step process.
First, you have to identify all of the mutually exclusive ways of partitioning the number $(6)$ into the sum of three non-negative integers.
These are
- 6,0,0
- 5,1,0
- 4,2,0
- 4,1,1
- 3,3,0
- 3,2,1
- 2,2,2
Then, one at a time, you have to enumerate each of the three mutually exclusive patterns above. In such an enumeration, the second step is to enumerate the total number of different ways of assigning suits, to fit the pattern. I refer to this as computing the suit-factor.
Then, the third step is to enumerate the number of ways of choosing the cards within each of the pertinent suits. I refer to this as computing the combination-factor.
Throughout this discussion, you need to know that if $~n \in \Bbb{Z^+},~$ and $~k \in \{0,1,2,\cdots,n\},~$ that the number of ways of choosing $k$ things out of $n$ things, sampling without replacement, where order of selection is regarded as irrelevant is
$$\binom{n}{k} = \frac{n!}{k![(n-k)!]}.$$
$\underline{\text{Case 1: Pattern = 6-0-0}}$
There are $(4)$ ways of choosing which suit to draw the cards from. So, you start with a suit-factor of $(6)$.
Then, you have the combination-factor, that represents the number of ways of selecting the cards, once the suit assignments are made. There are $~\displaystyle \binom{13}{6}~$ ways of choosing the $(6)$ cards within one suit.
Therefore, the overall enumeration here is
$$4 \times \binom{13}{6}. \tag1 $$
$\underline{\text{Case 2: Pattern = 5-1-0}}$
There are $(4)$ ways of choosing which suit to select $(5)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit to draw the single card.
So, the suit-factor is $(4 \times 3 = 12).$
Then, the combination-factor is
$~\displaystyle \binom{13}{5} \times \binom{13}{1}.$
Therefore, the overall enumeration here is
$$12 \times \binom{13}{5} \times \binom{13}{1}. \tag2 $$
$\underline{\text{Case 3: Pattern = 4-2-0}}$
There are $(4)$ ways of choosing which suit to select $(4)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit to select $(2)$ cards from.
So, the suit-factor is $(4 \times 3 = 12).$
Then, the combination-factor is
$~\displaystyle \binom{13}{4} \times \binom{13}{2}.$
Therefore, the overall enumeration here is
$$12 \times \binom{13}{4} \times \binom{13}{2}. \tag3 $$
$\underline{\text{Case 4: Pattern = 4-1-1}}$
There are $(4)$ ways of choosing which suit to select $(4)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit will be unused.
So, the suit-factor is $(4 \times 3 = 12).$
Then, the combination-factor is
$~\displaystyle \binom{13}{4} \times \binom{13}{1} \times \binom{13}{1}.$
Therefore, the overall enumeration here is
$$12 \times \binom{13}{4} \times \binom{13}{1} \times \binom{13}{1}. \tag4 $$
$\underline{\text{Case 5: Pattern = 3-3-0}}$
The suit-factor is $~\displaystyle \binom{4}{2} = 6.$
The combination-factor is $~\displaystyle \binom{13}{3} \times \binom{13}{3}.$
Therefore, the overall enumeration here is
$$6 \times \binom{13}{3} \times \binom{13}{3}. \tag5 $$
$\underline{\text{Case 6: Pattern = 3-2-1}}$
The suit-factor is $~\displaystyle 4 \times 3 \times 2 = 24.$
The combination-factor is $~\displaystyle \binom{13}{3} \times \binom{13}{2} \times \binom{13}{1}.$
Therefore, the overall enumeration here is
$$24 \times \binom{13}{3} \times \binom{13}{2} \times \binom{13}{1}. \tag6 $$
$\underline{\text{Case 7: Pattern = 2-2-2}}$
There are $(4)$ ways of selecting the suit to be unused. Therefore, the suit-factor is $4.$
The combination-factor is $~\displaystyle \binom{13}{2} \times \binom{13}{2} \times \binom{13}{2}.$
Therefore, the overall enumeration here is
$$4 \times \binom{13}{2} \times \binom{13}{2} \times \binom{13}{2}. \tag7 $$
$\underline{\text{Final Computation}}$
To get the overall enumeration, all that you have to do is use a hand calculator against the computations in each of
(1), (2), (3), (4), (5), (6), and (7), above.
Then, take the overall sum of these seven computations.
