Dividing to exclude order in counting I was studying up on counting. I came across this document, which says:

If you choose two things separately and multiply, your answer will include order. If you don't want that order, you either have to divide by an appropriate $r!$ or else find a different sequence of little choices.

Would someone be able to give me an example of this? I'm not understanding.
 A: Suppose you have a collections of some objects let us say you have five of them, say a bowl full of fruits, say an Apple, Banana, Orange, Pear, Lemon. 
If you now want to choose two of them: 


*

*for the first choice you have $5$ possibilities.

*for the second choice you have $4$ possibilities (one less as you have already chosen one fruit).
Now, you have $5$ choices and for each of the $5$ choices, you have $4$ choices. So, in total you have $5 \cdot 4 =20$ choices to first choose a certain fruit and then to choose a second fruit.
However, perhaps or even likely, you do not care so much if you take A first and then B or the other way round first B and then  A. Since you might only care that in the end you have an apple and a banana (the order is irrelevant for you). So, if the order is irrelevant for you theen first A and then B is essentially the same as first B and then A. 
Therefore you in fact only have $20/2 = 10$ choices. You need to divide by the ways to arrange $2$ distinct objects, which is $2! =2$.
Now if you choose $3$ fruits, you get $5\cdot 4 \cdot 3 = 60$ ways to choose a first fruit, a second fruit, a third fruit, but if you do not care about the order you need to divide by the number of ways to arrange $3$ distinct objects that is $3!=6$. Getting $60/6= 10$ possibilities.
A: I would like to count the number of ways to choose three crayons from the box of eight Crayolas.  There are 8 crayons to pick for my first choice, 7 left for the second choice, and 6 left for the last choice, for a total of $8\cdot 7 \cdot 6 = 336$.
However, 336 is only correct if I consider the  order of my  choices to be be important: Do I consider the choice of red, blue, green to be different from green, red, blue, or from red, green, blue?  Or are they all the same way of choosing three crayons?  Is the order important?
If the order is not important, then I have to divide by $3! = 6$ to get the correct number of choices, in which case there are only 56 possible choices of 3 crayons.
A: I was going to give you this example: The number of ordered pairs consisting of two distinct English letters is $26\times 25$, because you have $26$ options for the first letter and then $25$ for the second (only $25$, not $26$ because you're not supposed to repeat the first letter).  In the (somewhat sloppy) language you quoted, you choose the two letters separately and multiply the number of options, to get the number of ordered pairs.  If you want unordered pairs, i.e., just $2$-element sets of English letters, then the answer is only half that much, $26\times25/2$, because each $2$-element set, like $\{f,e\}$ comes from two ordered pairs, like $(f,e)$ and $(e,f)$.  More generally, to count $k$-element sets, you could first count ordered $k$-tuples of distinct objects and then divide by $k!$, because every $k$-element set comes from $k!$ ordered $k$-tuples (you can order the $k$ elements of the set in $k!$ different ways).
But then I looked at the notes you linked to, and I saw that this example is on page $1$ there. So probably you found it unsatisfactory and are looking for something else.  It would probably be useful if you explain the unsatisfactoriness, so that I or others can look for a more satisfactory answer.
