In a 52-deck card, draw 6. Probability of getting 3 different suits, an 8 and a face card? 
A standard deck of cards has 52 cards. If order does not matter, what are the chances of getting 3 cards of different suits, an 8 and a face card?

I thought like this:


*

*P(first card being a different suit) = $1/1 $

*P(second card being a different suit) = $39/51 $

*P(third card being a different suit) = $26/50$

*P(fourth card being an 8) = $4/49 $

*P(fifth card being a face card) = $12/48$
My answer : $\frac{1}{1} \cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{4}{49} \cdot \frac{12}{48} = 0.81\%$
 A: Total no of cases are ${52 \choose 6}$ , i am gonna write ${n \choose r }$ as nCr.
(I am assuming that we have to take in consideration that which suit are of the 8 and face cards are of).

Now as we need 6 cards of three 3 different suites, the possible sums can be:-
$a)1+1+4\ \ b)1+2+3\ \  c)1+1+1+3\ \ d)1+1+2+2\ \ e)2+2+2$

a)1+1+4 : rather than counting an 8 appears and an face cars appers it's easier to count total - either one of them does not appears
= Total - 8 does not appear - face card does not appear + both 8 and face card does not appear.
(13C1.13C1.13C4  -  12C1.12C1.12C4 - 10C1.10C1.10C4 + 9C1.9C19C4)$\times$(number of ways of selection of suits)
Selection of suits -> 4C2(for selecting 2 suits from 1 cards come in) .2C(1) (from where 4 cards come)=12
So it comes out 465132

Same way b)->2135520 ,c)->757672 ,d)->1820448, 
e)->570348

Final answer $\frac{465132+2135520+757672+1820448 +570348}{{52 \choose 6}}$
=$\frac{5749120}{20358520}$=0.28239380858
A: The chance of getting no $8$s is $\frac {48\cdot47\cdot46\cdot45\cdot44\cdot43}{52\cdot51\cdot50\cdot49\cdot48\cdot47}=\frac{48!46!}{52!42!}$
The chance of getting no face cards is $\frac{40!46!}{52!34!}$ 
The chance of getting neither an $8$ nor a face card is $\frac{36!46!}{52!30!}$ 
By inclusion-exclusion the chance of getting at least one $8$ and at least one face card is then
$$1-\frac{48!46!}{52!42!}-\frac{40!46!}{52!34!}+\frac{36!46!}{52!30!}=\frac{309821}{1017926}\approx 0.304365$$ 
It is not quite fair to say the chance of getting three suits is independent of the $8$ and face card probability because getting a lot of the same rank improves your chance for multiple suits and asking for an $8$ and a face card (very slightly) decreases the chance of lots of the same rank.  I am going to consider them independent.  
The chance of getting all six cards of one suit is 
$4\cdot \frac{13!46!}{52!7!}$  The chance of getting all six cards in two suits would be $6\cdot \frac {26!46!}{52!20!}$ but we have counted the all one suit cases twice each, so the chance of single suits three times, so we have to subtract it twice.  The chance of at least three suits is then 
$$1-6\cdot \frac {26!46!}{52!20!}+8\cdot \frac{13!46!}{52!7!}=\frac {365209}{391510}\approx 0.932822$$
The total probability is then $$\frac{113149417589}{398528208260}\approx 0.283918$$
Most of the risk of failure comes from not getting an $8$.
A: When order doesn't matter (i.e., the number of unordered samples/$\textit{combinations}$ is what we are interested in), then $n\choose{k}$ methods are usually the safest way to go.
With that said, when all outcomes of an experiment are equally likely, then the probability of an event $A$ is
$$
\mathbb{P}(A) = \frac{|A|}{|\Omega|},
$$
where $|\cdot|$ denotes the number of elements in the set, and $\Omega$ denotes the sample space.
For example, the probability of obtaining 2 aces, 1 King and a card not ace or King from a deal of 4 cards is
$$
\frac{{4\choose{2}} {4\choose{1}} {44\choose{1}}}{52\choose 4}.
$$
Think of it as "from the 4 aces available, we want 2" and so on. The handy thing about using $n\choose{k}$ approach is that once you've worked out what the correct numbers are, you don't have to worry about extra combinations on top of that.
So back to our example (I assumed from your working that we do not care what suit the 8 and the face card belong to). The number of ways 3 cards of different suits can be dealt from a pack is
$$
\text{(Number of ways of choosing 3 suits from 4)}\times\text{(Number of ways of choosing 1 card from each suit 3 times)}\\
= {4\choose{3}}{13\choose{1}}^3.
$$
With these 3 cards chosen, the number of ways of choosing one 8 from the four available is ${4\choose{1}}$; same for the King. This leaves us with
$$
\frac{|A|}{|\Omega|} = \frac{{4\choose{3}}{13\choose{1}}^3{4\choose{1}}{4\choose{1}}}{{52\choose{5}}}
\approx  0.054102 = 5.4102\%.
$$
I was concerned about the assumptions of the question, since it's not clear that the suit draws have to account for drawing an 8 or a King, but it may be the case that a priori we don't take this into account, since the cards are all being drawn simultaneously. I'm not sure. In any case, this would be my approach and I hope it at least sheds some light for you on how these questions can be tackled.
