Permutation Question with "ABERRATIONAL" - Get Help & Insight

[Stellaferox]

Hi,

I have this problem in which a word (lets say "ABERRATIONAL") is given. Now how many permutations (no unique words) can be made given that the 3 A's have to be next to each other?

My solution: regard the 3 A's as one block, so we have 12-3+1 = 10 letters. This gives 10! permutations. The 3 A's yield 3! permutations. Answer = 10!.3!

This is wrong. The answer should be 302.400.

Help and insight is appreciated!

regards,

Marc

 

[Jamma]

Your idea of counting the A's as one letter is the right way to approach this (you can just pretend that you throw them away any extra A's originally and start the problem from there).

Where you've gone wrong is considering two different orderings of the letters as necessarily being different. This matters because you have repetitions of other letters, such as the R. So you need to make sure that two permutations where all letters except the different R's are in the same place, for example, are considered as the same word.

e.g.AAABERRTIONL is one of your words, but in your scheme, if we permute the two R's this is a new word, when it isn't.

 

[Jamma]

Oh, sorry, is this what you meant by "non-unique words". Then I'm not sure what's wrong with your method.

 

[Stellaferox]

Coming from the dutch word "SINAASAPPELS" (oranges) I can understand that the answer should be 302.400 being 10!/(3!.2!) as the SSS and PP have to be counted as interchangable. But the question stated clearly permutaties and not combinations. So A1A2A3-bertionl is not the same as A3A2A1_berationl IMHO. Correct?

 

[Jamma]

If you are looking for all the words being unique and all the A's need to appear next to each other, my answer is 10!/2.

Where 302.400 comes from, I'm not sure. Are you sure you've read the question correctly?

 

[Jamma]

Do the R's have to go next to each other? Then I would get 9!.2!.3! (for the same reason as before, if the ordering of the R's matters).

So, if you want all unique words but where the ordering of the different letters matters, I would get:

12!=479001600

If the different orderings of the R's and A's doesn't matter so that I just want all words I can make with those letters, I get:

12!/(2!.3!)=39916800

If you want all words where the ordering of the letters matters and all the A's are next to each other, you get:

10!.3!=21772800

If you want all such words where the A's come next to each other, and the ordering of the same letters doesn't matter then you get:

10!/2=1814400

If you want all such words where the A's come next to each other, and the R's come next to each other, then you will get

9!=362880

and 9!.2=725760 if the order of the R's matters but still need to be next to each other.

None of these are low enough to be 302.400=120800, so I advise you look at the question again!

 

[Stellaferox]

No it's just the A's that have to be adjacent. I can see that the answer 302.400 for the dutch word is correct regarding just the unique words but the question was how many possibilities without regarding A1A2A3-bertionl and A2A3A1-bertionl as the same word.

 

[Jamma]

Well, then your original answer of 10!.3! is correct.

For simplicity, let's label the letters (except the A's) as x_1,x_2,x_3,...,x_n and A_1,A_2,...A_m for the A's, since these have special rules (they need to appear together).

Every word of x_i's and A_i's is uniquely determined by the instruction of a word made from A,x_1,x_2,...,x_n (where A denotes a block of A's) together with a permutation of the A_i's. This is quite easy to see (every such instruction defines a unique word and every word defines a unique such instruction).

So we have (n+1)!(m)! such expressions: the (n+1)! is where to place the n+1 terms x_1,x_2,...,A and the m! is the way of arranging the A_1,A_2,...,A_m.

In your case, the answer is 10!.3!. What you did initially is definitely correct, and 302.400 cannot be correct.

 

[Stellaferox]

Well, I thought so too. Thank you very much for your answer and time!

Marc