This is a follow up on my earlier question. To put some values and context, let me say I am trying to justify the number of possibilities of choosing 2 chapters out of 10 in a book i,e $10 \choose 2$. At this point this is just notation.
One way to enumerate the possibilities:
$\underbrace{\underbrace{(1,2),(1,3)\cdots,(1,10)}_{9}\underbrace{(2,1),(2,3)\cdots,(2,10)}_{9}\cdots\underbrace{(10,1),(10,2)\cdots,(10,9)}_{9}}_{10}$
So we have $10\cdot 9$ possibilities but we've double counted pairs such as $(n,m)$ and $(m,n)$ so we have $\frac{10\cdot 9}{2}$.
Now I try to reason with the sum and product rules I've learnt and once again Im confused, which might become detrimental if I go further.
Say I select chapter $1$. I can select chapter $2$ OR $3$ etc. Once I have selected one chapter, the selection of the other chapter is independent w.r.t one another, so using the sum rule I can make total $9$ possible selections once I have selected the first chapter. But then, I can make $10$ different choices of the first chapter. Now it is clear in my mind that each one of these ten possible selections come bundled with 9 possibilities, so multiply the two. But when I read the definition many times and try to bring in a "product-rule argument" and I just cannot. It should come as easily as the thought that I should multiply it, so I've only understood the product rule in words. Please enlighten me.
Another thing I noticed, was that this could be achieved using a purely sum rule argument. Consider the enumeration scheme where we avoid repetition altogether
$\underbrace{(1,2),(1,3),\cdots,(1,10)}_{9},\underbrace{(2,3),(2,4),\cdots,(2,10)}_{8}\underbrace{(3,4),(3,5),\cdots,(3,10)}_{7},\cdots\underbrace{(9,10)}_1$
$=9+8+\cdots+1=\frac{10\cdot 9}{2}$
Only the sum rule has been used in this argument. And it is much easier for me to understand. So, a parenthetical query: can a purely product rule argument be used for selections? It does not seem so for me. It appears the sum rule is fundamental and product rule is a special case of it. Which might seem like saying that multiplication is a special case of addition in discrete case. Once again, this last paragraph is just a parenthetical query and not my immediate concern at the moment.