The Number of Increasing Vectors $(x_1,...,x_k)$ Satisfying $1 \leq x_i \leq n$ and $x_1 < x_2 <...I want to find the number of increasing vectors $(x_1,...,x_k)$ satisfying $1 \leq x_i \leq n$ and $x_1 < x_2 <...<x_k$.

Examples of vectors satisfying these conditions
Let $n =5$ and $k=3$ 
$$(\_,\_,x_1,x_2,x_3)$$
$$(\_,x_1,\_,x_2,x_3)$$
$$(x_1,\_,x_2,\_,x_3)$$
$$(x_1,\_,x_2,x_3,\_)$$
where the spaces of the vector all numbered as follows, $$(1\,\,\,,2\,\,\,,3\,\,\,,4\,\,\,,5)$$
and in general $$(1,2,...,n)$$

My Approach
We will count the number of vectors $\underrightarrow{x}$ with components $\underrightarrow{x} = (x_1,x_2,...,x_k)$ with $ 1\leq x_i\leq n$.



*

*Implied Restriction On $x_1$


*

*The largest value $x_1$ can have is $n-k+1$.

*The smallest value $x_1$ can have is $1$.



*

*The Procedure
Count the number of different vectors $\underrightarrow{x}$ such that $\{x_1\in \mathbb{Z}: 1 \leq x_1 \leq n-k+1\}$
Each of those vectors $\underrightarrow{x}$ has $n-x_1\choose k-1$ different increasing vectors. That is, $n-x_1$ different positions available to the $k-1$ vectors.



*

*The Result
The number of increasing vectors $\underrightarrow{x}$ with $(x_1,...,x_k)$ satisfying $1 \leq x_i \leq n$ and $x_1 < x_2 <...<x_k$ is equal to 
$$\sum\limits_{i=1}^{t = n-k+1} {n-i\choose k-1} = {n-1\choose k-1} + {n-2\choose k-1}  + \cdot\cdot\cdot + {n-(n-k+1)\choose k-1}$$
or in a neater form, the total number of increasing vectors is equal to 
$$\sum\limits_{i=k-1}^{n-1} {i\choose k-1}$$

Edit After Reading More Material
$$\sum\limits_{i=k-1}^{n-1} {i\choose k-1} = \sum\limits_{i=k}^{n} {i-1\choose k-1} = {n\choose k}$$
This $$\sum\limits_{i=k}^{n} {i-1\choose k-1} = {n\choose k}$$ is knows as Fermat's Combinatorial Identity.

Please have a look at my solution and give any suggestions and hints you may have.
 A: The easiest thing to do is just note that any subset of size $k$ from $1,2,\ldots, n$ corresponds to an increasing sequence uniquely by just putting the elements in order. So you have ${n\choose k}$ such subsets, hence the same number of increasing sequences.
A: It is MUCH easier than all of that:
Choosing a vector with these conditions is the exactly the same than choosing a subset of the set $\{1,2,\ldots ,n\}$ with $k$ elements, because the only thing that you must decide for each element is whether it is in the vector/subset or it is not. The only condition is that you must choose $k$ elements.
So the answer is simply
$$\binom{n}{k}$$
Nevertheless I think that your approach is correct. And you have proved that
$$\sum_{i=k-1}^{n-1}\binom{i}{k-1}=\binom{n}{k}$$
which is a well-known formula.
In Tartaglia's (or Pascal's) triangle, that means that if you sum the $k-1$ first numbers of a NE-SW diagonal, the result is the number that is just at the right of the next number in that diagonal. See red and blue drawings in the figure.

