prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors

Question

prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors,so that all these factors are elements of $[k]$. Repetition of factors is allowed,

that is my attempt :

for explain my attempt let's take $n=7$ and $k=4$

so let's find the number of all product that have $7-4=3$ factors , so that all these factors are element of $[4]$

our concern is just finding the numbers of all these products , now we know $[4]=\{1,2,3,4\}$, but let's denote it by $\{e_1,e_2,e_3,e_4\}$,

note that $(e_k=k )$ $,k \in \mathbb{Z^+}$

we want to chose $3$ factors from $\{e_1,e_2,e_3,e_4\}$,(repetition of factors is allowed)

all ways to do that equal the number of all solutions of this equation $e_1+e_2+e_3+e_4=3$

for more explain let's take this solution $(2,1,0,0)$ that mean we choose this product $ 1.1.2$,

and if we take $(1,1,0,1)$ that mean we choose this product $1.2.4$

so the numbers of all products that have 3 factors from $[4]$equal the number of all solution of the above equation ,and that's equal ${7-4+4-1 \choose 4-1}$

and we can do the same thing with any $n$ and $k$

so finally products that have $n-k$ factors,so that all these factors are element of $[k]$ is a $n-k$-element multisets of the set $\{1,2,3,4,....k\}$,so the number of all this products is the number of all weak composition of $ n-k$ into $k$ parts and it is ${ n-k+k-1 \choose k-1}={ n-1 \choose k-1}$

does my attempt true?

Answer 1

Your identification of this problem with the number of solutions of $$\sum_1^k e_i=n-k$$ is problematical since you have not taken account of the fact that the same product can occur in different ways.

For example $$2\times4\times6=2\times3\times8.$$