We ought never to allow ourselves to be persuaded of the truth of anything unless on the evidence of our own reason – René Descartes

The Number of R-combinations Repetition Allowed

A note for myself.
The number of ways to choose r elements from a set of n elements with repetition allowed
can be calculated as the number of ways to arrange $n-1$ "|"s and r "x"s;
i.e. it can be shown that there exists a function from either set to another
with one-to-one correspondence so that both sets have
the same cardinal number by the definition of cardinality.
For example, the number of ways to choose three numbers out of $1,2,3,4,5$
is equal to the number of ways to arrange 5 - 1 = 4 "|"s and 3 "x"s;
e.g. xxx|||| means we choose three 1s and |xx||x| means we choose two 2s and one 4.
In this example, you can think of 4 + 3 = 7 positions to which each 4 "|"s and 3 "x"s will be assigned.

Since there are only two symbols "|" and "x",
once we have chosen one symbol the other one fills the remaining positions automatically.
Then, we have $r+(n-1)$ positions to which either n-1 "|"s or r "x"s are assigned;
i.e. we have $r+(n-1)$ positions from which to choose n-1 or r positions.
Thus, the number of ways to arrange "|" and "x" is $(\genfrac{}{}{0ex}{}{r+n-1}{r})=(\genfrac{}{}{0ex}{}{r+n-1}{n-1})$.