How many 6 digit numbers can you write? How many six-digit positive integers can you write, if each number must have strictly increasing digits from left to right
I thought it was 6! Butt hat is wrong.
How do you do these sort of problems using combinations he answer is:
$$\binom{9}{3}$$
How do they get that??
 A: 6! is all the way of ordering 6 distinct objects. It is not the correct answer for two reasons: one, which 6 objects are you considering? The digits 1 through 6? 3 through 9? Two, after specifying the objects, since you are considering all orderings (this is what ! does) you will get orderings which are not strictly increasing.
So how do they get $\binom{9}{3}$? Well first off notice since the problem specifies strictly increasing, no digits may repeat. Now we need to consider which 6 digits from 1 to 9 we will include in our number. $\binom{9}{6}$ represents all the ways  of selecting 6 distinct objects with no repeats from a pool of 9 distinct objects. Now we need to order them. But, given a set of distinct digits, theres only one way to arrange them in increasing order. Thus $1 * \binom{9}{6}$ is the total number of ways to select the type of number in question. Then note that $\binom{9}{6} = \binom{9}{3}$. This is because choosing 6 digits to include in your number is equivalent to choosing 3 digits to exclude. (in fact you could have started with $\binom{9}{3}$, but I think its more natural to consider which numbers are included first).
A: Another way to view this is to look at the increments from digit to digit. 
The minimum initial digit is (presumably) $1$, and the maximum possible final digit is $9$, so there are $8$ increments. If the first digit is not $1$ or the last digit is not $9$, we can imagine that the relevant increments happen "outside" the number, before or after.
Each of the first five numbers definitely has an increment attached to it, so there are $3$ more increments to insert into the sequence, and $7$  places to put them, so we can play "stars & bars" and add them in ${7+3-1 \choose 3} = {9 \choose 3}$ ways.
Note that for non-strictly increasing digits (ie. never-decreasing digits), there are no  "attached" increments any more and there would be ${7+8-1 \choose 8} = {14 \choose 8}$ options.
