Difficulty Understanding Application of the Multiplication Principle While leafing through "Introduction to Probability" (Hwang, Blitzstein), I encountered the following problem.

A standard deck of cards is shuffled well. Two cards are drawn randomly, one at a time without replacement. Let $A$ be the event that the first card is a heart, and $B$ be the event that the second card is red. Find $P(A|B)$.

I'm having difficulty formally determining $P(B)$. Intuitively, I know it's $\frac{1}{2}$, but I'm having difficulty formalizing my intuition.
Here's what the book says about $P(B)$,

$$P(B) = \frac{26\times 51}{52 \times 51} = \frac{1}{2}$$
since there are 26 favorable possibilities for the second card and for each of those the first card can be any other card.

I kind of understand this. The book says, "chronological order is not needed in the multiplication rule", but here I don't see exactly how the multiplication rule can even be used.
After all, order does seem to matter, since by the time the second card is being drawn there are only 51 possibilities not 52. So how can we apply the multiplication rule backwards here?
 A: Why can we apply the multiplication rule backwards?
The easiest way of thinking about it is that all of your events are of the form $(c_1,c_2)$ where the $c_i$ are cards and $c_1 \neq c_2$. 
For example $(5 \text{ hearts}, 10 \text{ diamond})$ is a valid event but $(5 \text{ hearts}, 5 \text{ hearts})$ is not. 
Computing the value of $p(B)$ is now equivalent to computing 

What is the number of events/draws $(c_1,c_2)$ such that $c_2$ is red?

Well for each $c_2$ for a red there is 51 possibilities for $c_1$
that satisfied the condition so that
\begin{equation}
N(B)= \sum_{c_2 \text{ is red}} \#(c_1 \neq c_2) = \sum_{c_2 \text{ is red}} 51 = 26 \cdot 51
\end{equation}
Now counting all of the possible draws we have that 
\begin{equation}
N(\text{All draws})= \sum_{c_2 } \#(c_1 \neq c_2) = \sum_{c_2 } 51 = 52 \cdot 51
\end{equation}
so that probability is 
\begin{equation}
P(B) = \frac{N(B)}{N(\text{All draws})}= \frac{26 \cdot 51}{52 \cdot 51} = 1/2
\end{equation}
Okay but, Why can we apply the multiplication rule backwards?
Well go through the whole argument and replace $c_2$ with $c_1$ and heart with red and it is still valid. 

Rigorously We have that we are actually considering all of the two subsets of the set $C$='cards"=$\{1,...,52\}$; so that for example counting all of the ways that you can get $\{x,2\}$ for any $x$ is the same as counting all of the ways that you can get $\{2,x\}$ for any $x$. 

Don't let the language confuse you; the most important strategy in proble-solving is to strip away the language of all the misleading details; i.e.

Find the right abstraction. Get rid of unnecessary details that cloud the mind and find good notation/definitions that make this possible.

I personally recommend Feller's introduction to probability theory; the whole first volume is discrete combinatorial probability theory and the book is considered a treasure amongst mathematicians.
