How to conceptualize "dividing out" a number (e.g. in permutations, Bayes' Theorem)? I'm trying to achieve a better conception of what it means to "divide out" a variable/number, because I'm currently have a lot of trouble justifying to myself why it actually works the way it does in certain contexts. I apologize if this is too much of an elementary question but I can't seem to come to a satisfying conclusion.
I understand division in the context of explicit numbers but here (see pg. 3) is an example of where I get confused:
"so the number of ways of selecting r objects and ignoring all permutations of the remaining $(n − r)$ objects not chosen is to divide out the unwanted permutations:
No. of perms. $= \dfrac{1 \times 2 \times 3 \times···\times (n − r) \times (n − r + 1) \times··· \times(n − 1) \times n}{1 \times 2 \times 3 \times··· \times (n − r)}$"
So basically, why does this work, and how would explain it to a clever 5 year old? 
Similarly, what does it mean to "divide out" $P(B)$ in the classic representation of Bayes' Theorem, i.e.
$P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}$?

Look at the number of permutations of $3$ things out of $5$.  Call the five things $ABCDE$.
$$
\begin{array}{c|c|ccc}
\overbrace{\begin{array}{cccc} ABC & ACB & ADB & AEB \\ ABD & ACD & ADC & AEC \\ \underbrace{ABE}_B & \underbrace{ACE}_C & \underbrace{ADE}_D & \underbrace{AED}_E \end{array}}^{\text{starting with $A$}} & \overbrace{\begin{array}{cccc} BAC & BCA & BDA & BEA \\ BAD & BCD & BDC & BEC \\ \underbrace{BAE}_A & \underbrace{BCE}_C & \underbrace{BDE}_D & \underbrace{BED}_E \end{array}}^{\text{starting with $B$}} & \cdots\cdots\cdots\cdots\cdots
\end{array}
$$
When three are listed in some order, the first one could be $A$ or $B$ or $C$ or $D$ or $E$.  Five choices.  Only two are listed above, but you can see what the other three are.  Within each of those, the second can be any of four.  And the third can be any of three.  So the total number of permutations in the list is $5\times 4\times3= 60$.
So what if we want combinations rather than permutations.  Notice that one of the permutations listed above is $BEC$.  In how many orders can $BEC$ be listed?  Here they are:
$$
\overbrace{BCE\quad BEC}^\text{starting with $B$} \quad \overbrace{CBE\quad CEB}^\text{starting with $C$} \quad \overbrace{EBC\quad ECB}^\text{starting with $E$}
$$
It is listed $3\times2\times1=6$ times.  That means that this combination got listed $6$ times in the first list above.  Every combination got listed $6$ times.   So divide the length of the first list by $6$:
$$
\frac{5\times4\times3}{3\times2\times1} = \frac{60}{6} = 10.
$$
In other words, the number of combinations is $10$.
