Probability of getting at least one spade from drawing 2 cards in succession 
Two cards are drawn in succession without replacement from a deck of 52 playing cards. Find the probability that at least one of them is a spade.

Here's my approach:
We can get a spade in $13 \choose 1$ ways, and the from the rest of the cards $(52-1 = 51)$, we can choose any card, so $52 \choose 1$. Thus, our number of favorable outcome would be ${13 \choose 1} {51 \choose 2} = 13\times51 = 663$, and the probability would then be $\frac{663}{1326}$
But, the book has a different answer, i.e.,it computes the number of cases where there is no spade and the number of ways for that is of course ${39 \choose 2} = 741$, and our required probability is then $1-\frac{741}{1326} = \frac{15}{34}$
Why are the two approaches different? Where am I going wrong?
 A: You've correctly identified that your sample space contains ${52 \choose 2} = 1326$ possible combinations of cards, but you are double-counting some of the combinations in your figure of $663$ ways of drawing at least one spade. Specifically, for every combination of two spades, you are counting it twice: if the first card is an ace of spades, and the second card a queen, it is counted separately from $(Q♠, A♠)$.
To fix this, you can just subtract the number of combinations of two spades. Then, $663 - {13 \choose 2} = 663 - 78 = 585$, and $\frac{585}{1326} = \frac{15}{34}$.
More generally, this results from the set "at least one spade" being the union of the sets "first card is a spade" and "second card is a spade".
The size of the union of two sets $S_1$ and $S_2$ is given by
$$|S_1 \cup S_2| = |S_1| + |S_2| - |S_1 \cap S_2|$$
This can be extended to a family of sets $\{S_i\}_{i \in I}$ for finite $I$:
$$\left| \bigcup_{i \in I}{S_i}\right| = \sum_{J \in \mathcal{P}(I)}{(-1)^{|J|-1}\left|\bigcap_{j\in J}{S_j}\right|}$$
A: An alternative rehabilitation of the approach taken by the OP is as follows:
You can either get:

*

*Spade, Non-spade : 
$\displaystyle \binom{13}{1} \times \binom{39}{1} = 507.$


*Spade, Spade : 
$\displaystyle \binom{13}{2} = 78.$
So, there are $(585)$ ways of drawing two cards, at least one of which is a spade.
So, the probability of drawing at least one spade is
$$\frac{585}{\binom{52}{2}} = \frac{585}{1326} = \frac{15}{34}.$$
A: Another way to describe the answer:
We can treat the order of draws as relevant and keep track of that in our numerator and denominator.  Break into cases based on when we first draw a spade.
We can get a spade on our first draw and then draw anything for our second card.
Or... we can draw a non-spade for our first card and then a spade on our second.
$$\dfrac{13\cdot 51 + 39\cdot 13}{52\cdot 51}$$
This, of course, equals the same as the other ways to present the answer.  The punchline is that there are some scenarios where treating order as relevant is useful, other scenarios where it is not.  You may choose whether you treat it as relevant or not.  It is your choice, and so long as you are consistent with your choice for both numerator and denominator and don't make mistakes then it shouldn't matter.
A: By designating a particular spade as the spade in your hand, you count each hand with two spades twice.
Let's do a direct count, then compare it with your answer.  There are $\binom{13}{1}\binom{39}{1}$ ways to select one spade and one card of a different suit.  There are $\binom{13}{2}$ ways to choose two spades.  Hence, the number of favorable cases is
$$\binom{13}{1}\binom{39}{1} + \binom{13}{2}$$
Since there are $\binom{52}{2}$ ways to select two cards from the deck, the probability of selecting at least one spade is
$$\frac{\dbinom{13}{1}\dbinom{39}{1} + \dbinom{13}{2}}{\dbinom{52}{2}}$$
By designating a particular spade as the spade in your hand, you count each hand with two spades twice, once for each way you could designate one of the spades as the spade in your hand.  For instance, if you draw $5\spadesuit, 7\spadesuit$, your method counts this hand twice:
$$
\begin{array}{c c}
\text{spade} & \text{additional card}\\ \hline
5\spadesuit & 7\spadesuit\\
7\spadesuit & 5\spadesuit
\end{array}
$$
Note that
$$\binom{13}{1}\binom{39}{1} + \color{red}{\binom{2}{1}}\binom{13}{2} = \color{red}{663}$$
