Soltaire card probability 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? What is your chance of winning?
 A: We first count the $3$-card sequences that have a $10$. The $10$ can be in any of $3$ places. For each choice of location for the $10$,  the leftmost vacant place  can be filled in $9$ ways. For each of these ways, the remaining vacant place can be filled in $8$ ways, for a total of $(3)(9)(8)$.
We now count the all odd sequences. The first number can be chosen in $5$ ways. For each such way, the second card can be chosen in $4$ ways, and then the third in $3$ ways, for a total of $(5)(4)(3)$. 
Thus our total is
$$(3)(9)(8)+(5)(4)(3).\tag{1}$$
As to your chances of winning, the total number of possible sequences is $(10)(9)(8)$, and they are all equally likely. Thus the probability of winning is the count in (1) divided by $(10)(9)(8)$. 
Remark: An alternate approach is to first count the number of winning hands (so we don't care about order). A hand that contains a $10$ can be chosen in $\binom{9}{2}$ ways, since we need to choose the remaining $2$ cards. A hand that is all odds can be chosen in $\binom{5}{3}$ ways. This gives a total of $\binom{9}{2}+\binom{5}{3}$. Multiply by $3!$ to take into account the fact that any hand can be arranged in $3!$ orders. 
A: We first count the 3-card sequences that if you don't win, namely not all of your cards are odd and you cannot use 10.
Thus the result is $$(\binom{4}{3}+\binom{4}{2}\binom{5}{1}+\binom{4}{1}\binom{5}{2})6=246.\tag{1}$$
(Note that we cannot use 10 so there are 4 even numbers to choose and because there are 6 different orders to arrange 3 cards.)
So the chance of winning is $720 - 444=276$.
($720=\binom{10}{3}6$.)
