How do I count the number of increasing functions from $\{1, 2, 7\}$ to $\{1, 2, 3, 4, 5, 6, 7, 8\}$? I can't seem to figure it out, one way I thought about it is:
Let's take $f(1) = 1$:  


*

*$f(2) = 1$ (because it's not strictly increasing) so for $f(7)$ I have $8$ possibilities  

*$f(2) = 2$ so $f(7)$ now has $7$ possibilities  

*$f(2) = 3$ so $f(7)$ now has $6$ possibilities
.
.
and so on. 


So for $f(1) = 1$ we have $8+7+6+5+...+1$ functions (this can be expressed as the formula $\frac{(n+1)n}{2}$)  
So for $f(1) = 2$:  


*

*$f(2) = 2$ results in $7$ possibilities of $f(7)$
.
.
and so on and it result in $7+6+5+..+1$  


so by my reasoning I would say that the number of increasing functions is equal to: $S_8 + S_7 + S_6 + ... + S_1$ (where $S_n = \frac{(n+1)n}{2}$ )  
But I'm pretty sure that's not right but I don't know how to think about it.
 A: A small variation of the theme. Note, that in order to determine the number of increasing functions the elements of the domain $\{1,2,7\}$ are not relevant. Just the number of elements $|\{1,2,7\}|=3$ of the domain is essential.

We can reformulate the problem: Find the number of triples $(x_1,x_2,x_3)$ of positive integers with
  \begin{align*}
  1\leq x_1\leq x_2\leq x_3\leq 8\tag{1}
  \end{align*}
This can be solved using stars and bars, here with $n=3$ and $k=8$ giving
  \begin{align*}
  \binom{n+k-1}{n}=\binom{10}{3}=120
  \end{align*}

A: The function $g$ given by $g(1)=f(1)$, $g(2)=f(2)+1$, $g(7)=f(7)+2$ is a strictly increasing map to $\{1,2,\ldots,10\}$ for every increasing $f$ and vice versa. The number of such $g$ is simly the number of ways to pick three distinct elements of a ten-element set, so $10\choose 3$.
A: First, to count the number of strictly increasing functions, note that any combination of $3$ numbers from the set $\{1,2,\ldots, 8\}$ gives such a function and all such functions correspond to a combination.  So there are $8 \choose 3$ strictly increasing functions.
Second, count the number of functions $f(x)$ where two numbers go to the same image and one is different.  Either $f(1)=f(2) \neq f(7)$ or $f(1)\neq f(2) =f(7)$.  These two cases are identical, so we count the first one and multiply by $2$.  By the same reasoning above, there are $2{8 \choose 2}$ of these functions.
Third, count the number of functions with $f(1)=f(2)=f(3)$.  There are $8$ of these.  
Final tally:  $${8 \choose 3} +2 {8 \choose 2} +8 = 120,$$
