Simplifying a Product

  • #1
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Homework Statement


This is a child thread I'm creating from a previous topic:
https://www.physicsforums.com/threads/combinatorics-problem.871661/#post-5473920

In that thread, I was helped to come up with the expression for the number of arrangements of R distinct types of objects given the number of objects for each object type: {r_i} I'm just trying to simplify it now. I wanted to double check the work. Something seems off.

Homework Equations


$$\prod {_{n - \sum{r_{i-1}}}C_{r_j}}$$
or more specifically,
$$\prod_{j=1}^{R} {_{n - \sum_{i=1}^{j}{r_{i-1}}}C_{r_j}}$$
$$r_0 = 0$$

The Attempt at a Solution


[/B]
$$_{m}C_{k} = \frac{m!}{k!(m-k)!}$$
$$\therefore \space\space\space\space\prod_{j=1}^{R} \frac{(n - \sum_{i=1}^{j}{r_{i-1}})!}{r_j!(n - \sum_{i=1}^{j}{r_{i-1}} - r_j)!}$$

$$\frac{(n - 0)! \cdot (n - r_1)! \cdot (n - r_1 - r_ 2)! ... }{(r_1! \cdot r_2! \cdot r_3! ... ) [(n - r_1)! \cdot (n - r_1 - r_2)! \cdot (n - r_1 - r_2 - r_3)! ... ] }$$

$$n!\prod_{j=1}^{R} \frac{1}{r_j!}$$


Let me know if I can clear anything up!
 

Answers and Replies

  • #2
andrewkirk
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Provided it is the case that ##\sum_{i=1}^Rr_i=n##, that looks correct to me.
 

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