How many $6$ digit numbers have their digits in increasing order? I can calculate the amount of ways you can choose $6$ digits out of $($1,2,3,4,5,6,7,8,9$)$, but this would include combinations where there are $2$ or more of the same digit. 
 A: Take the sequence of numbers $(1,2,3,4,5,6,7,8,9)$ and simply get rid of three of them.  How many ways can you do this?  It is equivalent to ask how many subsets of three elements you can take from a set of nine elements.  So the answer is $${9\choose 3}=84.$$
A: Answers to the following two questions should help.  
1.How many ways can you first choose 6 different digits from that set? 
2.Given that you have chosen 6 digits how many ways can you arrange them so they are in increasing order?
A: There is a fairly straightfoward way to find this out.
The key point is that there is only one possible arrangement in increasing order for $6$ given numbers.
This means that you just simply have to find out how many $6$ digit numbers are there.
Assuming that you want all of them (even with repeated digits),
there are $10$ choices for each digit. However, the first digit cannot be $0$.
Therefore, we have
$$9\times 10^5=900000$$
such possible numbers.
If you want only distinct digits, you can breakdown the question to choosing $6$ numbers from $10$ numbers. There are ${10\choose 6}=210$ ways to do so. However, $0$ cannot be in the first place.
So, we just subtract these cases.
There are ${9\choose 5}=126$ such cases.
Hence, the answer is
$$210-126=84$$
