smartbei

April 29th, 2007, 09:07 PM

Given a certain multiset ({a,a,a,a,a,a,a,a,a,a,b,b,b,b,b,b,b,b,b,b} for example) how do I find the number of possible permutations using all of the multiset's members?

I have read through the wikipedia articles on the matter (permutations, combinatorics, etc.) and am still somewhat lost. It is possible to actually find all the possible permutations, bu this is prohibitavely expenisve time-wise, one you go beyond 10-15 characters.

An equation, and/or an explanation will suffice.

Thanks for all replies!

yabbadabbadont

April 29th, 2007, 09:35 PM

Do your own homework... :D

cwaldbieser

April 29th, 2007, 09:43 PM

Given a certain multiset ({a,a,a,a,a,a,a,a,a,a,b,b,b,b,b,b,b,b,b,b} for example) how do I find the number of possible permutations using all of the multiset's members?

I have read through the wikipedia articles on the matter (permutations, combinatorics, etc.) and am still somewhat lost. It is possible to actually find all the possible permutations, bu this is prohibitavely expenisve time-wise, one you go beyond 10-15 characters.

An equation, and/or an explanation will suffice.

Thanks for all replies!

I think there would be 20!/(10! x 10!) permutations. That would be 184,756 if my arithmetic is correct.

I started with a simple case, and elaborated on it:

{a,b,c}

I know there are 6 permutations. I can write them all out or compute them as 3!.

{a,a,b}

I substitute a's for all the c's, and then cross out the duplicates. There are only 3 permutations. That makes sense to me, because where in the first example I could create more permutations by moving around the third element, I only have half as many elements now.

It looks like when there are duplicate elements, the number of permutations will be reduced. I was not sure what the divisor would be, but I am thinking it should be some function of the number of duplicates. I know with 4 distinct elements, there are 4! or 24 permutations.

With {a,a,a,b} there are only four permutations I can write out. This seems to fit 4!/(3! x 1!).

With {a,a,b,b} there are 6 permutations. This also fits 4!/(2! x 2!).

It is really just guesswork and trying to find a pattern rather than a more reasoned approach, but that is how I'd do it.

smartbei

April 29th, 2007, 10:38 PM

Heh - this isn't homework. I was intersted and couldn't find an amswer. :D

Thanks for the prompt response cwaldbieser!

Rumo

April 29th, 2007, 11:11 PM

20!/(10!*10!) is correct!

Here's a simple approach:

Suppose you have to pick 10 values out of 20 possible values. Each picked value now represents the position of an a (or b) in the succession of a's and b's.

You probably already noticed that this is equivalent to a simple lottery problem (with the solution n!/(k!*(n-k)!).

Ramses de Norre

April 30th, 2007, 12:30 PM

If you want a more precise explanation you might want to look for combinations and binomial coefficients, because that's what the n!/(k!*(n-k)!) is.

Ayman

April 30th, 2007, 02:08 PM

Actually, this is a multinomial coefficient (http://en.wikipedia.org/wiki/Multinomial_coefficient). Binomial coefficients (http://en.wikipedia.org/wiki/Binomial_coefficient) are used to calculate the number of distinct subsets that have k elements out of an n set, whereas multinomial coefficients represent the number of ways to permute n elements with k1 elements of type 1, k2 elements of type 2, ... etc

antiemptyv

July 20th, 2007, 07:14 AM

true. the multinomial theorem is a generalization of the binomial theorem.

Paul Miller

July 24th, 2007, 05:48 AM

Others have given you the answer and a pointer to the multinomial theorem, but here's an explanation of why it works.

Lets consider first the simpler case of a set of 20 distinct items. To determine the number of permutations of items in this set, we note that there are 20 choices for the first item, 19 for the second, and so on, for a total of 20!. If we were dealing with the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}, we would be done.

Now, let's say that 1-10 represent the letter 'a' and 11-20 represent the letter 'b', which is the example you gave. Let's start by labeling the a's and b's as follows:

\{ a_0, a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9,

b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9 \}

(The above is LaTeX formatting; the \'s are necessary to escape the curly brackets and the _1, _2, etc, denote subscripts.)

Now that we've attached subscripts to our a's and b's, we can distinguish them, so there are 20! arrangements as per previous logic. For the sake of simplicity, assume that the permutation we're looking at is the one in the code box above.

Now, if we take away the subscripts on the a's, what we have is this:

\{ a, a, a, a, a, a, a, a, a, a,

b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9 \}

The thing is, once we've removed the subscripts, we can't tell if the permutation we started with was our original one, or, say, this one

\{

a_9, a_8, a_7, a_6, a_5, a_4, a_3, a_2, a_1, a_0,

b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9 \}

It should be clear that once we remove the subscripts from the a's, we can't tell if we started with the first permutation of the subscripted letters or the second. In fact, we can't tell what order the a's were in at all, and there are 10! ways of putting them in order. Similarly, once we take the subscripts off the b's, there are 10! possible subscripted permutations we could have started with initially. Since the order of the a's and b's is independent, we multiply 10! x 10! to get the total number of unsubscripted permutations per subscripted permutation.

Thus, we end up with the result previously mentioned, that there are 20! / (10! x 10!) permutations of the 10 letter a's and 10 letter b's. This argument, in slightly more generality, is exactly the proof of the multinomial theorem.

Powered by vBulletin® Version 4.2.2 Copyright © 2017 vBulletin Solutions, Inc. All rights reserved.