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All coins of the same type are indistinguishable.

So I start with a 5 choose 3 for the pennies.

5!/(3!2!).

Giving me 10 ways to arrange the pennies.

For each combination of the pennies it seems there is only one way to place the two indistinguishable nickels. Is the final answer simply 10? That seems wrong.

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Yes, that is correct.

Even though there are $5!=120$ ways of physically doing it, many of them are equivalent (and you rightly corrected this by dividing by $3!2!$).

Look at them to convince yourself. The 10 distinct ways are listed below.

\begin{gather*}{\color{orange}P}{\color{orange}P}{\color{orange}P}{\color{gray}N}{\color{gray}N}\\ {\color{orange}P}{\color{orange}P}{\color{gray}N}{\color{orange}P}{\color{gray}N}\\ {\color{orange}P}{\color{orange}P}{\color{gray}N}{\color{gray}N}{\color{orange}P}\\ {\color{orange}P}{\color{gray}N}{\color{orange}P}{\color{orange}P}{\color{gray}N}\\ {\color{orange}P}{\color{gray}N}{\color{orange}P}{\color{gray}N}{\color{orange}P}\\ {\color{orange}P}{\color{gray}N}{\color{gray}N}{\color{orange}P}{\color{orange}P}\\ {\color{gray}N}{\color{orange}P}{\color{orange}P}{\color{orange}P}{\color{gray}N}\\ {\color{gray}N}{\color{orange}P}{\color{orange}P}{\color{gray}N}{\color{orange}P}\\ {\color{gray}N}{\color{orange}P}{\color{gray}N}{\color{orange}P}{\color{orange}P}\\ {\color{gray}N}{\color{gray}N}{\color{orange}P}{\color{orange}P}{\color{orange}P} \end{gather*}

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  • $\begingroup$ +1 for the colors — that looks great $\endgroup$ – Santana Afton Sep 11 at 23:52

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