Finding number of combinations So I'm trying to figure out how many 3-number combinations can be made in a specific range, but the combinations can ONLY be in increasing numeric value. So for example: 
If I get the range 1-50, a possible combination could be 1-24-44, but 45-23-27 is not a possible combination as it is not in increasing numeric order. You can't repeat numbers in the combinations.  
I don't need to show every single combination, just the number of combinations that can be made.
Any idea how I can do this? 
 A: Let's consider your example of selecting three numbers from the set $\{1, 2, 3, \ldots, 50\}$ that are in numerical order. There are $50$ ways to select the first number, $49$ ways to select the second number, and $48$ ways to select the third number.  However, the three numbers we have selected may not be in numerical order.  Let's say we have selected your example of $45$, $23$, and $27$.  You wish to reject this selection since they are not in numerical order.  Observe that there are $3! = 3 \cdot 2 \cdot 1 = 6$ sequences in which we could pick the subset $\{23, 27, 45\}$:
\begin{align*}
23, 27, 45\\
23, 45, 27\\
27, 23, 45\\
27, 45, 23\\
45, 23, 27\\
45, 27, 23
\end{align*}
This is because, assuming this subset was selected, there were three ways to pick the first number, two ways to pick the second number, and one way to pick the third number.  Thus, for a particular subset of three numbers, we will select them in the correct numerical order one sixth of the time.  Hence, we can select three numbers of the set $\{1, 2, 3, \ldots, 50\}$ in increasing numerical order in 
$$\frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1}$$ 
ways.  
Observe that if we multiply the numerator and denominator by $47!$, we obtain
$$\frac{50 \cdot 49 \cdot 48 \cdot 47!}{3 \cdot 2 \cdot 1 \cdot 47!} = \frac{50!}{3!47!} = \binom{50}{3}$$
which is equal to the number of three-elements subsets of $\{1, 2, 3, \ldots, 50\}$. 
Why should this be the case?  Observe that we could select a subset of three elements from the set $\{1, 2, 3, \ldots, 50\}$, then place the resulting elements in increasing numerical order.  There are $\binom{50}{3}$ ways to select a subset of three elements and just one way to place them in increasing numerical order. 
Hence, the number of ways to select three elements from a set of $n$ elements so that the elements are in increasing numerical order is equal to the number of three element subsets of an $n$ element set, which is $$\binom{n}{3} = \frac{n!}{3!(n - 3)!}$$
