Please refer to the following excerpt from "Pattern Recognition and Machine Learning" Bishop.

We can understand this alternative view of entropy by considering a set of N identical objects that are to be divided amongst a set of bins, such that there are $n_i$ objects in the $i^{th}$ bin. Consider the number of different ways of allocating the objects to the bins. There are N ways to choose the first object, $(N - 1)$ ways to choose the second object, and so on, leading to a total of N! ways to allocate all N objects to the bins, where N! (pronounced ‘factorial N ’) denotes the product N \times (N - 1) $\times \ldots \times 2 \times > 1$. However, we don’t wish to distinguish between rearrangements of objects within each bin. In the $i^{th}$ bin there are $n_{i}!$ ways of reordering the objects, and so the total number of ways of allocating the N objects to the bins is given by: $${\bf W} = \frac{N!}{\prod_{i}n_{i}!}$$ which is called multiplicity.

I am confused about the bold-faced part.

If bin 1 were to have n=1 object and bin 2 were to have 2 objects and so on, then the number of options for placing the objects in $i^{th}$ bin would be like - (with successive bin have only remaining objects to choose from):

$$^{N}C_{1} * ^{N-1}C_{2} * ^{N-3}C_{3}* ^{N-6}C_{4} \cdot \cdot $$

Were am I missing the point?

P.S.: If each bin has only one object, then I can understand N!.


1 Answer 1


You are not missing anything. You simply found another derivation of the same relationship.

We know that

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

If we expand the terms in your product,

$$^{N}C_{1} * ^{N-1}C_{2} * ^{N-3}C_{3}* ^{N-6}C_{4} \cdot \cdot $$

, we get

$$\frac{n!}{1!(n-1)!} \cdot \frac{(n-1)!}{2!(n-3)!} \cdot \frac{(n-3)!}{3!(n-6)!} \cdot \frac{(n-6)!}{4!(n-10)!} \dots $$

If we simplify the factorials,

$$\frac{n}{1!} \cdot \frac{(n-1)(n-2)}{2!} \cdot \frac{(n-3)(n-4)(n-5)}{3!} \cdot \frac{(n-6)(n-7)(n-8)(n-9)}{4!} \dots $$

And now it's obvious that this is

$$ \frac{n!}{\prod_i i!} $$

  • $\begingroup$ Got it! Thanks for clear, prompt response $\endgroup$ Commented Jun 5, 2019 at 9:49

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