Mathematical Permutations

[thanksforallthefish]

Given the following 7 letters

A A B C D E E

what is the maximum number of permutations using these letters??


My figuring says 3000 as there are

5 possibilities for the first letter A B C D or E .. (5)

assuming the first letter is an A
then there are 5 possibilities for the second letter A B C D or E .. (5 x 5)

assuming the second letter is an E
then there are 5 possibilities for the third letter A B C D or E ... (5 x 5 x 5)

then there are 4 possibilities for the fourth letter ... ( 5 x 5 x 5 x 4)


giving 5 x 5 x 5 x 4 x 3 x 2 x 1 = 3000 combinations

Is this correct ??? ... thanks

and am i correct in saying the minimum number would be 240?

 

[The Kraken]

Now do you want combinations or permutations?
Are the 2 A's and E's discrete entities?
Is the order of combination of importance?

the K

 

[thanksforallthefish]

Now do you want combinations or permutations?
Are the 2 A's and E's discrete entities?
Is the order of combination of importance?

the K

combinations or permutations ... damn ... not certain what the difference is

hmm ... i would like to know the maximum and minimum that use all of the letters, every one will have all 7 letters

the A's and E's are discrete so every one will have 2 A's and 2 E's

The order is not important.

Thanks

 

[The Kraken]

OK.

Your answer is:

nPr = n!/(n-r)! = 2580

the K

 

[thanksforallthefish]

Thanks ... however I cant seem to get the answer you do, would you be able to give an explanation?

nPr(n, r) The number of possibilities for choosing an ordered set of r objects (a permutation) from a total of n objects.
Definition: nPr(n,r) = n! / (n-r)!


I assume n = 7 and i guess r would also be 7 which would give 7! = 5040

this can't be right as there are only 5 different choices for the first letter ie A B C D or E

Thanks again

 

[The Kraken]

But you said that each of the A's and E's was a discrete entity.
Therefore one could consider one of the A's a "Z" and one of the E's as an "X" or whatever.

Ooops, my mistake.

Basically what you have is 7 factoral divided by 0 factoral which equals 1 which gives a result of 5040.

the K

 

[mikemill]

I'm betting that the letters will be used to create identifiers. As such the two As and the two Es are not discrete since (AA and AA produce the same identifier).

As such I would suspect the answer to be:
₅P₅ * 2 * 1 = 240

To get that you basically assume that the first (or last really doesn't matter) positions are determined by nPr. After that you have a choice of either A or E (* 2) and then you are left with the remaining character.

combinations or permutations ... damn ... not certain what the difference is

Permutations take into account position, so ABC is different from BAC.
Combinations only care about the occurrence of the items, so ABC is the same as BAC but different from BAD.

Poker hands work on combinations but license plates work on permutations.

 

[Abidon]

But you said that each of the A's and E's was a discrete entity.
Therefore one could consider one of the A's a "Z" and one of the E's as an "X" or whatever.

Ooops, my mistake.

Basically what you have is 7 factoral divided by 0 factoral which equals 1 which gives a result of 5040.

the K

:doctor: The funny thing is that I've never had to actually use that formula, or 99% of all the other formulas I learned, since I got my engineering degree 8 years ago. :dork:

 

[thanksforallthefish]

But you said that each of the A's and E's was a discrete entity.
Therefore one could consider one of the A's a "Z" and one of the E's as an "X" or whatever.

Ooops, my mistake.

Basically what you have is 7 factoral divided by 0 factoral which equals 1 which gives a result of 5040.

the K

I think the problem here is you have someone that thinks they know what they are talking about but actually don't ... namely ME

I'm betting that the letters will be used to create identifiers. As such the two As and the two Es are not discrete since (AA and AA produce the same identifier).

As such I would suspect the answer to be:
₅P₅ * 2 * 1 = 240

To get that you basically assume that the first (or last really doesn't matter) positions are determined by nPr. After that you have a choice of either A or E (* 2) and then you are left with the remaining character.


Permutations take into account position, so ABC is different from BAC.
Combinations only care about the occurrence of the items, so ABC is the same as BAC but different from BAD.

Poker hands work on combinations but license plates work on permutations.

So what I am after is unique permutations, but each one must contain all 7 letters

I think I know what I want now


Thanks for the help guys.

 

[spoolin01]

I'm going to make a leap of intuition, as opposed to reasoning it through carefully, and say the answer is

7!/4

The clue came from Mike Mill's example of the indistinguishability of AA and AA. For every pattern of As there is a complete set of duplicates, so reduce 7! by a factor of 2. Take that half set and do the same for the Es and you get 7!/4.

That's my line for now.