Am I doing this probability problem correctly? You are playing a solitaire game in which you are dealt three cards without replacement from a simplified deck of $10$ cards (marked $1$ through $10$). You win if one of your cards is a $10$ or if all of your cards are odd. How many winning hands are there if different orders are different hands?


*

*$P(\text{1 card is a 10}) = 9 \times 8$ because if one of them is a 10, there are 9 options for position 2, and 8 for position 3. Multiply this by 3 because the order changing counts as another hand. (specified). $P(\text{All Odd}) = 5 \times 4 \times 3$ because there are $5$ and then there are $4$ and so on, multiplied by $3$ for the same reason.

*How many winning hands...$396$, adding those two together.
The next question is chance of winning. The total amount of possibilities would be $10!$ wouldn't it? So $396/10!$ ? It's been a while since I took combinatorics and this is a preliminary review/diagnostic for a CS course, and I feel like with a little review I know more than might be apparent without it.
 A: Your first part produces $216$, which is correct.  We can verify this by finding the total number of hands, and subtracting the number of hands that are drawn from just the cards $1$ through $9$:
$$
_{10}P_3\, - \,_9P_3 = \frac{10!}{7!} - \frac{9!}{6!} = 720-504 = 216
$$
The second part is not correct; it should be just the number of three-card permutations drawn from the set of five odd-numbered cards:
$$
_5P_3 = \frac{5!}{2!} = 60
$$
You can't reorder them (and multiply by $3$) because you've already counted them all in multiplying $5 \times 4 \times 3$: Each reordering corresponds to a hand already counted.
No hand can both contain a $10$ and contain only odd-numbered cards, so the total number of winning hands is just $216+60 = 276$.  The total number of hands, as previously alluded to, is the number of three-card permutations drawn from the entire deck of ten cards:
$$
_{10}P_3 = \frac{10!}{3!} = 720
$$
So the desired probability is
$$
P(\text{win}) = \frac{276}{720} = \frac{23}{60}
$$
A: I am led to believe that order does not matter. These kinds of problems are good to be dealt with in the following manner.
$(1):$ Calculate the number of possible outcomes.
This is simply the number of possible opening opening hands. That's $\binom{10}{3}=120$.
$(2):$ Calculate the number of success cases.
For a $10$ in hand, there are $\binom{9}{2}$ possibilities: $1$ of the cards is a $10$, the others are any two cards from the remaining nine.
For an all-odd hand, there are $\binom{5}{3}$ possibilities: any three of the five odd cards.
Since these success cases are mutually exclusive, we simply add them up, for a total of $\binom{9}{2}+\binom{5}{3}=36+10=46$ success cases.
$(3):$ Divide one by the other.
The probability is hence $\frac{46}{120}=\frac{23}{60}\simeq38.34\%$.
