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Re: k-permutations enumeration
*To*: mathgroup at smc.vnet.net
*Subject*: [mg116376] Re: k-permutations enumeration
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Sat, 12 Feb 2011 05:20:00 -0500 (EST)
*References*: <ij0eaq$9eo$1@smc.vnet.net> <201102110914.EAA07802@smc.vnet.net> <4D55A228.1070908@wolfram.com>
Daniel Lichtblau wrote:
> Ray Koopman wrote:
>> On Feb 10, 2:23 am, Michal Kvasnicka <michal.kvasni... at gmail.com>
>> wrote:
>>> How can I find the number of k-permutations of n objects, where there
>>> are x types of objects, and r1, r2, r3 ... rx give the number of each
>>> type of object?
>>>
>>> Example:
>>> I have 20 letters from the alphabet. There are some duplicates - 4 of
>>> them are a, 5 of them are b, 8 of them are c, and 3 are d. How many
>>> unique 15-letter permutations can I make?
>>>
>>> In the example:
>>>
>>> n = 20
>>> k = 15
>>> x = 4
>>> r1 = 4, r2 = 5, r3 = 8, r4 = 3
>>>
>>> Furthermore, if there isn't a straightforward solution: how
>>> efficiently can this problem be solved? Any Mathematica code?
>>
>> Please disregard my previous post, which counted only
>> "words" containing all of the x different "letters".
>>
>> k = 15; r = {4,5,8,3};
>> Tr[Multinomial@@@Select[Tuples[Range[0,#]&/@r],Tr@#==k&]]
>>
>> 187957770
>
> Another approach is to use an exponential generating function. It seems
> to be fairly fast.
>
> multvals = {4, 5, 8, 3};
> select = 15;
> factors = Map[summand[x, #] &, multvals];
> Coefficient[Factorial[select]*Times @@ factors, x^select]
>
> 187957770
>
> Daniel Lichtblau
> Wolfram Research
Ray Koopman pointed out to me that I forgot to show the definition of
summand. (He even provided me with a shorter variant, which I will use.
How thoughtful.)
summand[x_,n_] := Sum[x^r/r!,{r,0,n}]
Daniel Lichtblau
Wolfram Research
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