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Topic: unique ordering
Replies: 9   Last Post: May 10, 2014 1:16 PM

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Jussi Piitulainen

Posts: 317
Registered: 12/12/04
Re: unique ordering
Posted: May 10, 2014 1:16 PM
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Math Lover writes:

> In article <0acdace5-c53a-4ca4-a474-200d90f2ef4f@googlegroups.com>,
> <radams2000@gmail.com> wrote:
>

> > Thought experiment:
> >
> > Lets say that I randomly draw N letters out of a Scrabble
> > letter-bag. I will then place those letters in a horizontal line
> > on a table. How many UNIQUE patterns of letters can I produce? If
> > there were no duplicates it would be easy (N factorial) but when
> > there are duplicate letters it seems more complicated.

>
> Let's start with two letters, you have "m" of one and "n" of the
> other. So the total number to place them is (m+n)!, but you
> have to divide by m! to account for the fact that all different
> orderings of the "m's" yield the same word, and you have to also
> divide by n! for the same reason; so it's
>
> (m+n)!
> ------
> m!n!
>
> You can also think about it as choosing "m" place for the first
> letters and "n" for the other, so it's clearly (n+m choose n).
>
> The general case is exactly the same.


The letters were drawn from a bag.

If there are m letters of the first kind and n letters of the second
kind in the bag, and no other kinds, and both m > 1 and n > 1, there
are four different 2-sequences: aa, ab, ba, bb.

The number of different 3-sequences depends on whether m = 2 or n = 2,
since aaa or bbb may or may not be available.

It's easy in terms of the multiplicities among the letters in hand
after the draw but seemingly not in terms of the multiplicities in the
bag before the draw.



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