Combinatorics of the word RAKSH

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The discussion revolves around calculating the number of groups formed from the letters in "RAKSH" that contain at least one of each letter when selecting from n slips. It is noted that groups of 1 to 4 slips are irrelevant since there are 5 distinct letters. The initial approach suggests fixing the letters R, A, K, S, and H, and then filling the remaining slips with any of the five letters. Participants express confusion about the problem's clarity and seek assistance in determining the probability of selecting k slips that include all five letters. The conversation highlights the intersection of combinatorics and probability, with some participants indicating they have not yet studied probability.
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Homework Statement


There is a word given:
"RAKSH"
and n slips are provided. A person is free to write anyone of the letters (R,A,K,S,H) in each of the slips. Repetition is allowed, i.e. for eg. one such case would be that all the 'n' slips are filled with the letter "R'.

Then we begin our task:
First we groups of 1 slip from n
then groups of 2 slips from n
then groups of3
4,5,6,7...n.

Find the number of such groups formed that contain at least one of each of the letters, i.e. R,A,K,S,H!

The Attempt at a Solution



I was told that its a difficult question. Here is what I think:
it is obvious that groups of 1 to 4 members are useless. since there are 5 letters in RAKSH.

First I consider those cases in which at least one of each letter is there:
5 slips have been fixed as RAKSH. and there are remaining n-5 slips , each have 5 options to get filled with.
so is the answer 5n-5?
help me!
 
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The statement of the problem is not completely clear to me. Is this correct?

There are n slips of paper. Each slip is printed with a letter. For each slip, the letter has been selected at random, with equal probability, from a list of five letters.

We now choose k slips at random. What is the probability that each of the five letters occurs at least once?
 


Avodyne said:
The statement of the problem is not completely clear to me. Is this correct?

There are n slips of paper. Each slip is printed with a letter. For each slip, the letter has been selected at random, with equal probability, from a list of five letters.

We now choose k slips at random. What is the probability that each of the five letters occurs at least once?

Yup!
 


I am waiting for some help!
 


Well, if that's the setup, then the total number n of slips doesn't matter. Each of the k slips is equally likely to have any letter on it.

Can you write down the probability that there are exactly n_R slips with R, n_A slips with A, etc?
 


Avodyne said:
Well, if that's the setup, then the total number n of slips doesn't matter. Each of the k slips is equally likely to have any letter on it.

Can you write down the probability that there are exactly n_R slips with R, n_A slips with A, etc?

I am studying Permutation an Combination. I haven't yet studied probability.
 

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