What does $\binom{a}{b}$ represent? Problem:
How many six-digit positive integers can you write, if each number must have strictly increasing digits from left to right
From the other link, 
How do I know if I use $\binom{a}{b}$ or factorials?
What does $\binom{a}{b}$ represent? And when should I use it?
Thanks you!
 A: In rough words, it represents a way of selecting members of  a group, such that the order doesn't matter. 
$$\binom{a}{b} = \frac{a (a - 1) \cdots (a - b + 1)}{b(b-1)\cdots 1}= \frac{a!}{b! (a-b)!} $$
Edit: Adding an answer to your problem, from $9$ numbers you choose $6$. As you want them in a strictly increasing order, once you've picked them there is only $1$ way to set them, so the possibilities are $$\binom{9}{6} \dot \ 1 = \binom{9}{6}$$  
A: $a \choose b$, spelled "$a$ choose $b$", is the coefficient of $x^b$ in the binomial expansion of $(x + 1)^a$; thus called "binomial coefficient" also. It's usually useful when you are randomly selecting a combination of $b$ objects (taken at once), out of a collection of $a$ objects. It's necessary that $b \le a$. It gives you the number of possible such combinations.
A: Hopefully I haven't made a mistake in the following.
Imagine three red $r_1,r_2,r_3$and five blue beads $b_1,b_2,b_3,b_4,b_5$ in a bag.  
Suppose you select 3 balls at random (one at a time) and put them into boxes.  
The first box can be filled in 8 ways.
The next box can be filled in 7 ways.
The last box can be filled in 6 ways.
In total there are $8\times7\times6$ ways of doing this if we consider $r_1,r_2,b_3$ different to $r_2,r_1,b_3$ ie not just two red then a blue.
Notice that $8\times7\times6=\frac{8\times7\times6\times5\times4\times3\times3\times2\times1}{5\times4\times3\times3\times2\times1}=\frac{8!}{(8-3)!}$ due to (a lot) of cancelleing.
This is called $^8P_3$
But we might want to see $r_2,r_1,b_3$ as $r,r,b$ ie not be able to tell the difference between $r_1$ and $r_2,$ for example.  But for any section of three there'd be $3!$ ways they could look the same.  If this is the case we have $3!$ too many and actually want $\frac{8!}{3!(8-3)!}$
This is sometimes called $^8C_3=\binom{8}{3}$
