Combinatorics of the word RAKSH

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Homework Help Overview

The problem involves combinatorial reasoning related to the word "RAKSH" and the selection of letters from it. Participants are tasked with determining the number of groups formed from n slips that contain at least one of each letter from the set {R, A, K, S, H}.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • One participant suggests that groups of 1 to 4 members are not useful since there are 5 letters. Another participant questions the clarity of the problem statement and seeks confirmation on the setup regarding the selection of letters and slips.

Discussion Status

There is ongoing exploration of the problem's setup and assumptions. Some participants are attempting to clarify the conditions under which the groups are formed, while others are considering the implications of the total number of slips on the probability of selecting each letter at least once.

Contextual Notes

Participants note that the problem statement may not be fully clear, particularly regarding the random selection of letters and the relevance of the total number of slips.

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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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