Does $\frac{(x + n-1)!}{(x-1)!n!}$ hold?

Question

I made an equation that calculates all of the possible combination of products of primes in which the equation looks like this $\frac{(x + n-1)!}{(x-1)!n!}$. $x$ represents the number of the primes we're going to use

Note ($1$ is also included) and $n$ represent the empty slots for the primes.
Example:
$n = 2, x = 2 ;$
then
$(2, 2), (2,1), (1,1)$

Note: $(1,2)$ and $(2,1)$ are the same thing because $2\times 1=2$ and $1\times 2 = 2$. Using the equation we get $3$ as the answer which is correct. To generalize, if we were to just plugin $n = 2$ we will have the following equation $\frac{(n(n+1))}{2}$. I wonder if for any n $\in$ $\Bbb Z^+$ the equation still holds. Is there a way to prove whether or not it's true or false and if so then it would be great if you can share it.

Answer 1

${n \choose m} = \frac {n!}{m!(n-m)!}$ is always an integer.

And your expression is ${x+n-1\choose n} = \frac {(x+n-1)!}{n!([x+n-1]-n)!} =\frac {(x+n-1)!}{(x-1)!n!}$

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Postscript:

(${n \choose m} = \frac {n!}{m!(n-m)!}= \frac {\prod_{k=1}^m (n-m + k)}{m!}=\frac {\text{product of m consecutive integers}}{\text{product of each integer up to m}}$)

(As $m$ consecutive integers will contain a multiple of every integer up to $m$, each integer up to $m$ will divide one or more of the terms. [If two or more integers divide the same term then the greatest common divisor of the two integers will divide another nearby term and the product of the integers will divide the product of the terms.] The product of the terms will contain multiples of each integer up to $m$ and be divisible by $m!$[1]).

(Not to mention.... if you count the number of ways to choose $m$ elements from $n$ objects there must be a whole number of ways to do it...[either you choose $m$ objects or you don't; there's no ways to have a partial choice]... and our calculations do assure it can be calculated as $\frac {n!}{m!(n-m)!}$. so that must be an integer.)

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[1][This is a rough argument... To formalize we have to do an induction that to prove $7!$ divides $9\cdot 10\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15$ that $4,6$ each divide a term, but they both divide the same term, $12$, but that is okay.. As $6|12$ then $6|\prod 9...15$ and as $4|12$ so $4|\prod 9....15$ and so $\operatorname{lcm}(4,6)|\prod 9...15$ and $4\cdot 6|(\prod 9....15)\cdot \gcd(4,6)$. But as $\gcd(4,6) < 7$ then we must have $\gcd(4,6)|12 \pm \gcd(4,6)$ and $9 \le 12\pm \gcd(4,6) \le 15$ so $4\cdot 6| 12 \cdot [12 \pm \gcd(4,6)] $ and $4\cdot 6| \prod 9.... 15$

[This leads to a lot of infinite regress shunting and some hard to put into words induction but ... it all works out.

[$1|11; 2|10; 3|15; 4|12; 5|10$ and yes we had $2|10$ as well but $2, 5$ are relatively prime so $2\cdot 5| 10$. and $6|12$ and $\gcd(4,6)=2|14$. And $7|14$ and although we already had $\gcd(4,6)=2|14$ we have $2$ and $7$ are relatively prime so $\gcd(4,6)\cdot 7|14$. And so...... $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7|9\cdot 10\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15$]

(Heres a better more thorough argument provided by Mike Earnest Why is $(n+1)(n+2)\cdots(n+s)$ divisible by $s!$? )