What is the relation between permutation and combination in ${}_nC_k$? Suppose we flip 3 coins. The possible outcomes can be pictured, with heads in black and the number of heads denoted $k$, as:
⚪⚪⚪ $k=0$
⚪⚪⚫ $k=1$
⚪⚫⚪ $k=1$
⚫⚪⚪ $k=1$
⚪⚫⚫ $k=2$
⚫⚪⚫ $k=2$
⚫⚫⚪ $k=2$
⚫⚫⚫ $k=3$
In the binomial probability formula ${}_n C_k p^k q^{n-k}$, where $n$ is the number of coins, the binomial coefficient is
$${}_n C_k = {n \choose k} = \frac{n!}{k! (n-k)!} $$
and counts the number of ways we can choose $k$ items out of $n$.
We can also represent the factorials in this formula using the physical coins, so long as we number or colour them. Here we have:
$n! = 3! = 6$ because
❶ ❷ ❸
❶ ❸ ❷
❷ ❶ ❸
❷ ❸ ❶
❸ ❶ ❷
❸ ❷ ❶
If we are interested in getting two heads, so $k = 2$, then
$k! = 2! = 2$ because
❶ ❷
❷ ❶
How can we picture the relationship between the case $k=2$ in our shaded/unshaded diagram, which shows there are 3 ways:
⚪⚫⚫
⚫⚪⚫
⚫⚫⚪
with the calculation derived from the numbered/coloured diagrams, dividing the 6 ways ❶ ❷ ❸, ❶ ❸ ❷, ❷ ❶ ❸, ❷ ❸ ❶, ❸ ❶ ❷, ❸ ❷ ❶ by the 2 ways ❶ ❷, ❷ ❶ to also make 3?

I suspect it is more like this:

*

*Colour the $n$ coins with $n$ colours (one each) in any of $n!$ ways.


*Change a specific $k$ of the colours to black - these $k$ could have been ordered in any of $k!$ different ways but we want to treat them as the same order


*Similarly change the other $n-k$ colours to white - these could have been ordered in any of $(n-k)!$ different ways but we again want to treat them as the same order


*So the number of ordering $n$ coins with $k$ black and $n-k$ is $\,_nC_k=\dfrac{n!}{k!\, (n-k)!}$
and an attempt at a diagram for $\,_3C_2=\dfrac{3!}{2!\, 1!}=\dfrac{6}{2}=3$ might be more like this:

