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?

• This is an example of a widespread general phenomenon I describe at stats.stackexchange.com/a/288198/919.
– whuber
Commented Feb 4, 2023 at 23:02
• I have substantially rewritten for clarity in the hope this question can be reopened - both @whuber and Henry seem to have understood the point (hopefully I have as well!!) but if I've somewhat spoiled the question in some way, please feel free to re-edit Commented Feb 18, 2023 at 15:42

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:

• Thank you very much! Perfect clear, totally understandable! Commented Feb 6, 2023 at 15:02