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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?

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    $\begingroup$ This is an example of a widespread general phenomenon I describe at stats.stackexchange.com/a/288198/919. $\endgroup$
    – whuber
    Commented Feb 4, 2023 at 23:02
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    $\begingroup$ 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 $\endgroup$
    – Silverfish
    Commented Feb 18, 2023 at 15:42

1 Answer 1

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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:

enter image description here

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  • $\begingroup$ Thank you very much! Perfect clear, totally understandable! $\endgroup$
    – RodParedes
    Commented Feb 6, 2023 at 15:02

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