The number of $n$-tuples with a sum less than $m$ is $C_m^n$? I accidentally found that the number of positive integer $n$-tuples that has its sum no greater than $m$ is $m$-choose-$n$ ($C_m^n$). For example, when $n=3$ and $m=4$ we have the following tuples that satisfy:
$(1,1,1)\quad (1,1,2)\quad (1,2,1)\quad (2,1,1),$
whose number is exactly $C_4^3=4$.
But I failed to see why. Any hints?
 A: The number of solutions in the positive integers of the inequality
$$x_1 + x_2 + \cdots + x_n < m$$
is the number of solutions in the positive integers of the equation
$$x_1 + x_2 + \cdots + x_n + x_{n + 1} = m\tag{1}$$
where $x_{n + 1} = m - (x_1 + x_2 + \cdots + x_n)$.  The number of solutions of equation 1 is the number of ways $m$ ones can be separated into $n + 1$ nonempty blocks of ones by placing $n$ addition signs in the $m - 1$ spaces between the ones, which is 
$$\binom{m - 1}{n}$$
The number of solutions in the positive integers of the equation
$$x_1 + x_2 + \cdots + x_n = m$$
is the number of ways $m$ ones can be separated into $n$ nonempty blocks of ones by placing $n - 1$ addition signs in the $m - 1$ spaces between the ones, which is 
$$\binom{m - 1}{n - 1}$$
Hence, the number of solutions in the positive integers of the inequality
$$x_1 + x_2 + \cdots + x_n \leq m$$
is 
$$\binom{m - 1}{n} + \binom{m - 1}{n - 1} = \binom{m}{n}$$
by Pascal's Identity.
A: Suppose you are on the real line, you are in zero and you can just do $n$ steps and will go no further than $m$.
$$0---1---2---...m-2---m-1---m $$
Then the number of $n$-uples of positive integers whose sum is less than $m$ is exactly the number of way to  go from $0$ to somewhere before $m$. That is to $(k_1,...,k_n)$ you associate the following "path" :
$$(k_1,k_1+k_2,...,k_1+...+k_n) $$
Then, what you get is actually a subset $A$ of $\{1,...,m\}$ of cardinal $n$. This is an easy (I leave it to you) verification that every subset of cardinal $n$ gives such a $n$-uples and that those two functions are inverse.
But now we know the number of subsets of $\{1,...,m\}$ which are of cardinal $n$...
A: We can imagine that we have to assign $r$ balls to $n$ boxes for each possible $r$ with $n+r\leq m$. It is just counting the number of $0$-$1$-strings with $r$ zeros (balls) and $n-1$ ones (fences), thus there are $C_{n-1+r}^r$ combinations. Hence, in total, $C_{n-1+0}^0+C_{n-1+1}^1+C_{n-1+2}^2+\dots+C_{n-1+m-n}^{m-n}=C_m^{m-n}=C_m^n$.
