Deck of Cards Probability Dividing by n! 1) Assume we have a standard poker deck of 52 cards, 13 each with 4 suits. What is the probability of picking 3 same numbered cards and 2 same numbered cards (A triple and a pair so AAABB).
Solution: (13C1)(4C3)(12C1)(4C2)/(52C5)
2) Let's say you have a deck of cards with only a 10, Jack, Queen, King and Ace of which the cards only belong to the suits of Heart and Diamonds. So there are 10 cards in total. Of the 10 cards you deal yourself a hand of 5 random cards. What is the probability of drawing no pairs??
You select one of the 5 pairs (5C1) and of that 2 pair you select one random car (2C1)
Now you have 4 remaining pairs. You pick a pair (4C1) and from that pair you select a card (2C1)
...
So you have
(5C1)(2C1)(4C1)(2C1)(3C1)(2C1)(2C1)(2C1)(1C1)(2C1)=5!*2^5

we divide by 5! because of duplicates (order does not matter) then 
divide by the total number of possible combinations (10C5)
Ans: 
2^5/(10C5)

Now this is my question: why is it in the first problem we did not divide by n! and in the second problem we did?? You divide by n! when order doesn't matter. But why did we not take that into account in the first problem but did do so in the second?
 A: For the first question, it is probably a typo, but the answer is not correct. 
There are $\binom{13}{1}$ ways to choose the "number" you will have $3$ of. The rest is as you have it. The total number of "full house" hands is
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}.$$
I would do the second question differently, though with the same final answer. There is only one way, or if you prefer $\binom{5}{5}$ ways, to select the kinds of cards in your hand. Now for each kind, you have $2$ choices of actual card, for a total of $2^5$. Then we divide by $\dbinom{10}{5}$.
Your way is fine too. You imagined picking the cards in order, one at a time, and divided by $5!$ to compensate for the multiple counting. Instead, for the denominator you could also consider ordered choices of $5$ cars, of which there are $(10)(9)(8)(7)(6)$. 
The reason we did not have to consider order in the first problem is that for a full house, you choose what kind you will have $3$ of, and what kind you will have $2$ of. If we were counting $2$ pair hands (so $2$ pairs and an odd card), saying we can choose the "first" (??) kind in $\binom{13}{1}$ ways and the second in $\binom{12}{1}$ ways would give a double-counting error: we then have to divide by $2!$. Or alternately one can say we can choose the $2$ kinds that we have a pair in in $\binom{13}{2}$ ways.
