Number of Compositions

Theorem
A $k$-composition of a non-negative integer $$n$$ is an ordered $k$-tuple: $$c = \left({c_1, c_2, \ldots, c_k}\right)$$ such that $$c_1 + c_2 + \cdots + c_k = n$$ and $$c_i $$ are strictly positive integers.

The number of $$k$$-compositions of $$n$$ is $$\binom{n-1}{k-1}$$ and the total number of compositions of $$n$$ is $$2^{n-1}$$ (i.e. for $$k = 1, 2, 3, \ldots, n$$).

Proof
Consider the following array consisting of $$n$$ ones and $$n-1$$ blanks:

$$\begin{bmatrix} 1 \ \_ \ 1 \ \_ \ \cdots \ \_ \ 1 \ \_ \ 1 \end{bmatrix}$$

In each blank we can either put a comma or a plus sign.

Each way of choosing $$,$$ or $$+$$ will give a composition of $$n$$ with the commas separating the individual $$c_i$$'s.

It follows easily that there are $$2^{n-1}$$ ways of doing this, since there are two choices for each of $$n-1$$ blanks and the multiplication rule then gives the result.

Similarly if we want specifically $$k$$ different $$c_i$$'s then we are left with choosing $$k-1$$ out of $$n-1$$ blanks to place the $$k-1$$ commas.

The number of ways of doing so is $$\binom{n-1}{k-1}$$ from the Binomial Theorem.