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I have a keypad lock with digits 0-9. It takes an $n$ digit keycode to successfully open, with repetitions allowed. Magically, let's say I know every distinct digit that goes into the correct keycode. There are $m$ such digits. How many possible combinations are there with those digits?

Here's what I have so far. Let's say there are n distinct digits in the keycode (i.e., $m$ = $n$). The answer then is, trivially, $n!$.

If $m = n-1$, we know that one of the digits appears twice. So, we have the number of combinations as, $\frac{m \cdot n!}{2!}$.

Is there a general formula for $(m,n)$?

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    Just working this out. Once n is greater then m then the extra digits can come as partition of (n-m). – 2017-02-13
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    Sorry got m and n mixed up. – 2017-02-13
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    http://mathworld.wolfram.com/Multichoose.html – 2017-02-13
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    @spaceisdarkgreen could you explain a bit further? I don't see how to mix in permutations with the multichoose function, especially when different combinations have different number of permutations... – 2017-02-13
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    @Naumz You're right, it's more complicated cause you have different numbers of permutations for each choice – 2017-02-14

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There are $n$ empty positions in which we can put our available digits

$$\underbrace{\text{_ _ _ _ ... _ _}}_{n\text{ positions}}$$

each successive position may take on any of the $m$ possible digits hence there are

$$\text{possible key codes}=m^n\qquad\blacksquare$$

However, this counts keycodes length $n$ that use at most $m$ digits.

In order to count those keycodes that use at least one of each digit form sets $A_k$ of keycodes that use only the $m$ known digits but don't use the digit $k$ (which is amongst those $m$ digits). Then there are $\binom{m}{1}$ sets $A_k$ with cardinality:

$$|A_k|=(m-1)^n$$

because there are $(m-1)^n$ keycodes using the remaining $m-1$ digits. Similarly there are $\binom{m}{2}$ set intersections (keycodes that don't use digits $k$ and $l$) which have cardinality:

$$|A_k\cap A_l|= (m-2)^n$$

this time there are $m-2$ digits for each position of the the keycode. higher order intersections look similar, for the $\binom{m}{3}$ 3-intersections we have:

$$|A_k\cap A_l\cap A_p|=(m-3)^n\\\vdots$$

Using the principle of inclusion-exclusion the number of valid keycodes that belong to none of the $A_k$ is given by:

$$\boxed{\text{valid keycodes}=m^n-\binom{m}{1}(m-1)^n+\binom{m}{2}(m-2)^n-\binom{m}{3}(m-3)^n+\ldots +(-1)^{m-1}\binom{m}{m-1}1^n\qquad\blacksquare}$$

This is related to the Stirling Numbers of the second kind $S(n,m)$ via

$$\text{valid keycodes}=m!S(n,m)$$

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    Perhaps your $m$ and $n$ are switched? The answer does not seem to match the solution for the two base cases in the question – 2017-02-24
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    Well, I checked using the example of $n=5$ for the two base cases and verified $5!=5^5-\binom{5}{1}4^5+\binom{5}{2}3^5-\binom{5}{3}2^5+\binom{5}{4}1^5$ and $\frac{4\cdot 5!}{2!}=4^5-\binom{4}{1}3^5+\binom{4}{2}2^5-\binom{4}{3}1^5$ . Also checked for some other values of $n$ and they all agree with your base cases. – 2017-02-24
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    oh, my apologies. I interpreted your answer incorrectly. Would you mind enclosing the answer in a box? I'll accept it. Thank you! – 2017-02-24