${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$ i tried to prove this by using the concept of composition
${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$
my attempt:
for explain my idea let's take $ n=6$ and $k =4$
so ${ 6 \choose 4}={ 5 \choose 3}+{ 4 \choose 3}+{ 3 \choose 3}$
${ 6 \choose 4}$is the number of composition of $7$ into $5$ blocks, so it is the number of solution of this equation $a+x+y+z+w=7$ when $x,y,z,w,a \in \mathbb{Z^{*+}}$
let's denote the number of solutions  of equation $A$ by $N_s(A)$
so now ,it's easy to see that $N_s(a+x+y+z+w=7)$ =$N_s(x+y+z+w=6)$ + $N_s(x+y+z+w=5)$ +$N_s(x+y+z+w=4)$
so it is ${ 5 \choose 3}+{ 4 \choose 3}+{ 3 \choose 3}$, and that is the RHS
and we can do that with any $n$ and $k$
so the LHS ${ n \choose k}$ = $N_s (\sum_{i=1}^{k+1} x_i=n+1)$
and the RHS ${ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$ = $N_s (\sum_{i=1}^{k} x_i=n)+N_s (\sum_{i=1}^{k} x_i=n-1)+N_s (\sum_{i=1}^{k} x_i=n-2)+...+N_s (\sum_{i=1}^{k} x_i=k)$
and because $N_s (\sum_{i=1}^{k+1} x_i=n+1)=N_s (\sum_{i=1}^{k} x_i=n)+N_s (\sum_{i=1}^{k} x_i=n-1)+N_s (\sum_{i=1}^{k} x_i=n-2)+...+N_s (\sum_{i=1}^{k} x_i=k)$
we can say that :
${ n \choose k}={ n-1 \choose k-1}+{ n-2 \choose k-1}+{ n-3 \choose k-1}+...+{ k-1 \choose k-1}$
Is my attempt correct?
 A: Yes, your proof is valid.  In words, you counted the number of solutions of the equation
$$x_1 + x_2 + \cdots + x_{k + 1} = n + 1 \tag{1}$$
in two different ways.  The left-hand side (LHS) counts the number of solutions directly.  The right-hand side (RHS) counts the same thing, depending on the value of $x_{k + 1}$ since you added the number of solutions of each of the equations
$$x_1 + x_2 + \cdots + x_k = n + 1 - x_{k + 1}$$
as $x_{k + 1}$ varies from $1$ to $n + 1 - k$, the set of all possible values $x_{k + 1}$ can assume if equation 1 is an equation in the positive integers.
Alternate Proof:  We count $k$-element subsets of the $n$-element set $$\{0, 1, 2, \ldots, n - 1\}$$
in two different ways.  The LHS gives a direct count.  The RHS counts the number of $k$-element subsets whose largest element is $j$ since if $j$ is the largest element, we must also select $k - 1$ of the $j$ elements smaller than $j$ to be in the $k$-element subset.  Since the largest element can vary from $k - 1$ to $n - 1$, we obtain
$$\binom{n}{k} = \sum_{j = k - 1}^{n - 1} \binom{j}{k - 1} = \binom{k - 1}{k - 1} + \cdots + \binom{n - 3}{k - 1} + \binom{n - 2}{k - 1} + \binom{n - 1}{k - 1}$$
