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I have $1,2,\ldots, n$ numbers and I want pick $k$ of them with replacement and such that order matters.

So for $n=10$ and $k=4$ I can get: $(1,2,2,4), (1,2,4,2), (1,2,3,10)$,...etc

I then have $n^k$ possible combinations. But now I only want to count the tuples which have a unique number. So $(1,2,2,4)$ and $(1,1,1,2)$ would be included but $(1,1,2,2)$ would not be included since both 1 and 2 are not unique? How do I count these?

I figured that I can pick a number out of $n$ for the first element in my tuple and then the remaining $k-1$ elements out of the remaining $n-1$ numbers, so the number of combinations would be $n\,(n-1)^{k-1}$. Since I have $k$ possible locations for the unique number I get $k\, n\, (n-1)^{k-1}$. However, clearly I am counting some combinations multiple times and I am not sure how to discount them.

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You can do this using inclusion–exclusion.

There are $k$ conditions, one for each of the $k$ positions containing a unique number. Any $j$ particular conditions can be fulfilled in

$$ \frac{n!}{(n-j)!}(n-j)^{k-j} $$

different ways (which for $j=1$ is your expression for one particular condition). Thus, by inclusion–exclusion the number of ways to fulfill at least one of the conditions is

$$ \sum_{j=1}^k(-1)^{j+1}\binom kj\frac{n!}{(n-j)!}(n-j)^{k-j}\;. $$

For $n=10$ and $k=4$, this is

$$ \sum_{j=1}^4(-1)^{j+1}\binom4j\frac{10!}{(10-j)!}(10-j)^{4-j}=9720\;. $$

Of course, for this particular small case we could have counted more easily that there are $\binom42\binom{10}2=270$ tuples with two pairs and $10$ with four identical entries, for a count of $10^4-270-10=9720$.

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  • $\begingroup$ Thanks. I didn't know about the inclusion-exclusion formula. Could you please expand on the first expression for the fulfilment of $j$ conditions? How did you arrive to that expression? $\endgroup$ – Tohiko Apr 16 '20 at 12:14
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    $\begingroup$ @Tohiko: To fulfill $j$ particular conditions, we need $j$ unique numbers. The first can have any of $n$ different values, the next any of the remaining $n-1$ different values, and so on, so there are $\frac{n!}{(n-j)!}$ ways to choose the unique numbers. The remaining $k-j$ numbers can each be any of the remaining $n-j$ numbers, so there are $(n-j)^{k-j}$ ways to choose them. $\endgroup$ – joriki Apr 16 '20 at 12:31
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For $j=1,\dots,n$ let $A_{j}$ denote the set of tuples that have $j$ as unique number.

Then to be found is $\left|A_{1}\cup\cdots\cup A_{n}\right|$ and for this we can use inclusion/exclusion and symmetry: $$\left|A_{1}\cup\cdots\cup A_{n}\right|=\sum_{j=1}^{n}\binom{n}{j}\left(-1\right)^{j-1}\left|A_{1}\cap\cdots\cap A_{j}\right|$$

Observe that for $j>k$ term $\left|A_{1}\cap\cdots\cap A_{j}\right|$ takes value $0$ so that these terms can be left out.

Further we have: $$\left|A_{1}\cap\cdots\cap A_{j}\right|=\frac{k!}{\left(k-j\right)!}\left(n-j\right)^{k-j}$$ resulting in: $$\sum_{j=1}^{\min\left(n,k\right)}\binom{n}{j}\left(-1\right)^{j-1}\frac{k!}{\left(k-j\right)!}\left(n-j\right)^{k-j}$$

Also observe that: $$\binom{n}{j}\frac{k!}{(k-j)!}=\binom{k}{j}\frac{n!}{(n-j)!}$$showing that this corresponds with the answer of Joriki.

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