# Talk:Ordered partitions

(Redirected from Talk:Ordered Partitions)

## (turns out not) incorrect math statement

The task says:

- Note that the number of elements in the list is

But this cannot be right.

The task with args 1,2,4 would generate 105 distinct partitions. But `4 choose 1`

is 0 and `4 choose 2`

is 0 and `4 choose 4`

is 1, so the above formula would give a result of 6. --Rdm 22:27, 7 February 2011 (UTC)

- Never mind, I fixed it. --Rdm 22:36, 7 February 2011 (UTC)

- Is it really fixed? As far as I know choose, the bigger number should be on the left. Anyways, I'll change it to

- to make it clearer --Eugen 10:36, 8 February 2011 (UTC)

- Oh! No, the original page was right, and I was wrong, about that left vs. right thing with the word "choose".

- My problems were twofold: First, I did not have a definition of "choose", and second I did not know how to group operations when faced with a + b choose c. I went with a + (b choose c), but it's now clear to me that I should have gone with (a+b) choose c. The current main page shows operator precedence clearly, but still could do with a link to a definition. (The link to Combinations suggests a definition, and there can be good reasons to order the arguments that way, but that does not actually define the operation the original (anonymous) author intended for
`choose`

. Perhaps the right approach here would be to mention popular notation issues on the Combinations page?) --Rdm 15:55, 8 February 2011 (UTC)

- My problems were twofold: First, I did not have a definition of "choose", and second I did not know how to group operations when faced with a + b choose c. I went with a + (b choose c), but it's now clear to me that I should have gone with (a+b) choose c. The current main page shows operator precedence clearly, but still could do with a link to a definition. (The link to Combinations suggests a definition, and there can be good reasons to order the arguments that way, but that does not actually define the operation the original (anonymous) author intended for

- (Note: I was the anonymous user that created the page, only after I created it I registered). You are right, that the first definition was ambiguous so I improved it. I also provided links to Ordered Partitions and Combinations for definitions of the binomial coefficient.

- If I understand your point about notation issues correctly you mean that the notation on Combinations is different from the notation of Ordered Partitions and that is why popular notation issues should be added to the Combinations page? Actually, I don't know if there is a standard notation for the
*enumeration of combinations*. There is the binomial coefficient notation for the*number of combinations*, though, and this is why I used it on the page Ordered Partitions. I think the Combinations is clear enough imho. --Eugen 11:18, 9 February 2011 (UTC)

- If I understand your point about notation issues correctly you mean that the notation on Combinations is different from the notation of Ordered Partitions and that is why popular notation issues should be added to the Combinations page? Actually, I don't know if there is a standard notation for the

- Fair enough. Thank you. --13:15, 9 February 2011 (UTC)

## Discarded code

I wrote this code for Ruby. It seems to return the correct result, but the algorithm seems to be too slow, so I will not post it on the page. --Kernigh 04:13, 10 February 2011 (UTC)

<lang ruby>require 'set'

def partitions(*args)

(1..args.reduce(:+)).to_a.permutation.inject(Set[]) do |set, perm| set << (args.inject([]) do |dist, arg| dist.push(perm.shift(arg).to_set) end) end

end</lang>