Probability that exactly k of N people matched their hats [SRoss P63 Ex 2g] 
The match problem stated in Example 5m of Chapter 2 (of A First Course in Pr, 8th Ed, Ross) showed that the probability of no matches when $N$ people randomly select from among their own $N$ hats $= P[N]= \sum_{0 \le i \le N}(-1)^i/i!$
  What is the probability that exactly $k$ of the $N$ people have matches?
Solution: Let us fix our attention on a particular set of $k$ people and determine the
  probability that these $k$ individuals have matches and no one else does. Letting $E$
  denote the event that everyone in this set has a match, and letting $G$ be the event that
  none of the other $N − k$ people have a match, we have
  $P(E \cap G) = P(E)P(G|E)$ 
  (Rest of solution pretermitted)
$P(E) = \dfrac{\text{ 1 choice for C1 } \times ... \times \text{ 1 choice for C(k - 1) } \times \text{ 1 choice for C(k) } \times N - k \text{ choices for C(N - k) }\times \ N - k - 1 \text{ choices for C(N - k - 1)} \times ...}{N!}$
  , where $C(k) =$ chap $k$, chaps 1 through k each has one choice due to their success in finding their hat, and the $P(E \cap G) = P(E)\binom{N}{k}P[N - k]$. 

I see that $P(E) \neq P(E \cap G)$, but I don't apprehend the method and still deem $G$ redundant.
Since $E$ is the event that exactly these $k$ people, for some $k$, have a match,
how and why isn't the required probability just $P(E)$? 
Since there are only $N$ people, thus the occurrence of $E$ (coincidently, directly, and straightaway) equals the occurrence of $G$?    


Since $E$ is the event that exactly these $k$ people, for some $k$, have a
  match, how and why isn't the required probability?

Because 


*

*You are asked for the probability that exactly $k$ people (no more and less) match their hats. The event that the (say) first $k$ match is not necessarily a "success", because there can be more matches in the remaining people.

*Furthermore, the event $E$ is the probability that a particular set of $k$ people matches
