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For a set of $N$ total objects where there can be repeats (multiple indistinguishable copies of an object), i.e., objects $X_1,...X_m$ such that $|X_1| + ... + |X_m| = N$, the number of distinct distinguishable permutations is given by $$\frac{N!}{|X_1|!*...*|X_m|!}$$ (see distinct permutations using all objects with repeated objects)

Is the number not starting with a given object $X_i$ (i.e, when you write down the permutation, the first object is not a $X_i$) then

$$\frac{N!}{|X_1|!*...*|X_m|!} - \frac{N!}{|X_1|!*...*|X_m|!}*\frac{|X_i|}{N}$$ ?

Tommy
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  • Your edit looks correct. The expression after the equal sign is the same, after canceling, as the expression in my answer below. – G Tony Jacobs May 16 '17 at 15:59
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    Note that your answer simplifies to $\frac{(N-|X_i|)(N-1)!}{|X_1|! \cdots |X_m|!}$, which can be obtained by first counting all permutations which do not start with an object of type $X_i$ (including all copies), of which there are $(N-|X_i|)(N-1)!$, and then dividing by $|X_1|! \cdots |X_m|!$ to account for the fact that all permutations of the $|X_k|$ objects of type $X_k$, $k=1,\dots,m$, are indistinguishable. – kccu May 16 '17 at 16:02

1 Answers1

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If you want to exclude the permutations starting with an object from $X_i$, just count up how many permutations there are of that description, and subtract from the total.

The quantity you need to subtract should be something like:

$$\frac{(N-1)!}{|X_1|!*\cdots*|X_i-1|!\cdots*|X_m|!},$$

because you're basically placing an $X_i$ object first, and then counting ways to arrange the remaining $N-1$ objects.

Does that make sense?

G Tony Jacobs
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