On permutations and combinations and probability From an ordinary deck of playing cards, cards are to be drawn successively at random without replacement. What is the probability of having two spades in first five draws ? 
I know this is a trivial question but I got little confused 
Actually I found this question in introduction to mathematical statistics 7th edition by Hogg and Craig on page 22 example 4.3
My intuition says I should use permutations (as we are drawing cards successively - - not simultaneously) to compute this, but the author is using combinations to compute this probability ..
Kindly give some suggestions. Where I am wrong in thinking about this .. 
Thankyou in advance . And sorry for asking a trivial question like this. 
 A: Case 1) If there are at least 2 spades in the first 5 cards
As for the sample space there are $$\binom {52}{5}.5!$$ ways to select and then draw the cards without any restriction.
We first find the number of ways in which there are exactly 2 spades in the first 5. Then we find the number of ways in which there are exactly 3 spades out of 5 and so on.
For two exactly two spades using the reasoning in case 2, we get the number of ways for two spades as $$\binom {13}{2}\binom {39}{3}.5!$$
Using this same reasoning we find answer for 3 spades, 4 spades and 5 spades. Hence we need to simply find $$\sum_{k=2}^5 \binom {13}{k}\binom {39}{5-k}5!$$
Which by Vandermonde's identity simplifies to 
$$5!\sum_{k=2}^5 \binom {13}{k}\binom {39}{5-k}=5!*\left[\binom {52}{5}- \left[ \binom {39}{5}+\binom {13}{1}\binom {39}{4}\right]\right]$$
Hence probability is $$1- \frac {\binom {39}{5}+\binom {13}{1}\binom {39}{4} }{\binom {52}{5}}$$
Case 2) If exactly 2 spades in first 5 cards
As for the sample space there are $$\binom {52}{5}.5!$$ ways to select and then draw the cards without any restriction.
But for our event we need two of the first 5 cards to be spade. We can select this spade in $\binom {13}{2}$ ways while the rest 3 can be chosen in $\binom {39}{3}$ ways. 
But since the cards are drawn successively drawn we also need to arrange them
Hence number of ways of  drawing 5 cards such that 2 of them are spade is $$\binom {13}{2}\binom {39}{3}.5!$$
Hence the probability is $$\frac {\binom {13}{2}\binom {39}{3}}{\binom {52}{5}}$$
A: For this problem, combinations is more natural.

Even though the cards are not dealt simultaneously, for the event to happen, the only thing that matters is what $5$-card subset you get, not the order in which the cards are dealt.

The number of $5$-card subsets is ${\large{\binom{52}{5}}}$.

Of those, assuming you want exactly two spades, you need a $2$-card subset of spades, and a $3$-card subset of non-spades.

The number of $2$-card subsets of spades is ${\large{\binom{13}{2}}}$.

The number of $3$-card subsets of non-spades is ${\large{\binom{39}{3}}}$.

Thus, by the multiplication rule, there are ${\large{\binom{13}{2}\binom{39}{3}}}$ qualifying hands, so the probability is
$$\frac{{\large{\binom{13}{2}}}{\large{\binom{39}{3}}}}{{\large{\binom{52}{5}}}}=\frac{9139}{33320}\approx .2742797119$$

If instead of exactly $2$ spades, you want at least $2$ spades, you can add the probabilities of exactly $k$ spades, for $k=0$, and $k=1$, and then subtract from $1$, so the probability would be
$$
{\large{1-\frac{\binom{13}{0}\binom{39}{5}+\binom{13}{1}\binom{39}{4}}{\binom{52}{5}}}}=\frac{1223}{3332}\approx .3670468187
$$
A: Posting as an answer instead of a comment:

Well, there usually is more than one way to solve a math problem.
For example, you can calculate the sample space of this problem as either $P_{52, 5}$ or $ C_{52, 5} \cdot 5!$
The number of ways of choosing $5$ cards in order (permutations) is the same as the number of ways of choosing $5$ cards without order (combinations) multiplied by the ways of ordering $5$ cards.
So, whether to use combinations, permutations, or both depends on the question and your approach.
