Permutation Combination - Different interpretations ?

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Hi All

This question can be solved in 2 ways depending on interpretation.

Question:
There is a set of characters, A B C D E F G H I J K. There are 4 character and 3 character codes to be made out of these characters. What is the ratio of the total number possibilities of the 4 character codes to the total possibilities of the 3 character codes?

I can solve the question.But the question is , in this question do we assume that total no of characters = 11 or can these alphabets be repeated??

Which interpretation would be correct.

I think bcos a set of chars is given and bcos we should not assume anything, there are only 11 characters and they cannot be repeated.

Any thoughts??
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by sanju09 » Wed Mar 28, 2012 11:48 pm
rainbownlife wrote:Hi All

This question can be solved in 2 ways depending on interpretation.

Question:
There is a set of characters, A B C D E F G H I J K. There are 4 character and 3 character codes to be made out of these characters. What is the ratio of the total number possibilities of the 4 character codes to the total possibilities of the 3 character codes?

I can solve the question.But the question is , in this question do we assume that total no of characters = 11 or can these alphabets be repeated??

Which interpretation would be correct.

I think bcos a set of chars is given and bcos we should not assume anything, there are only 11 characters and they cannot be repeated.

Any thoughts??
The shortcoming is with the question, my dear friend. Let's take both cases one by one.

Case 1: No letter is repeated.

The total number possibilities of the 4 character codes = 11P4 = 11! /7!

The total number possibilities of the 3 character codes = 11P3 = 11! /8!

Required ratio = (11! /7!)/ (11! /8!) = 8:1.

Case 2: Letters can be repeated to any possible number of times.

The total number possibilities of the 4 character codes = 11^4

The total number possibilities of the 3 character codes = 11^3

Required ratio = 11^4/11^3 = 11:1.
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by [email protected] » Thu Mar 29, 2012 11:06 pm
Yes sanju09 is perfectly correct. You will always get 2 cases in here...

and one more point is that, always keep permutation and combination as far as possible.

This is a permutation question and so no combination formula required. That is how it is...


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