4-letter words can be made from pulleys?

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Homework Help Overview

The discussion revolves around the combinatorial problem of determining how many 4-letter words can be formed from the letters of the word "PULLEYS." The problem involves considerations of letter repetition and arrangement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different cases for forming words, including arrangements with all different letters and those with repeated letters. Questions arise regarding the correctness of the original poster's calculations and the teacher's feedback.

Discussion Status

The discussion is ongoing, with participants sharing their interpretations and calculations. Some express confidence in their answers, while others question the assumptions made in the original solution. There is no clear consensus yet, as differing opinions and interpretations are being explored.

Contextual Notes

Participants note the constraints of the problem, including the specific letters available and the implications of letter repetition. There is mention of communication challenges with the teacher regarding the correct answer.

James_fl
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Hi, could someone please verify my answer? Thanks.. :smile:


How many 4-letter words can be made from the letters of the word PULLEYS? Explain your answer.

There are two mutually exclusive cases by which four letter-word can be arranged from the word PULLEYS. Let n(A) be the number of arrangements of 4-letter word with all different letters. Let n(B) be the number of arrangements that contain two L's and two other letters.

Case A:
There are C(6, 4) ways to choose subsets of four different letters. Each of these subsets has the length of four and can generate 4! sequences of letters. Therefore, n(A) = C(6, 4) x 4! = 360.

Case B:
There are C(2, 2) X C(5, 2) subsets that contain two L's and two other letters. The letters in each subset can be arranged in: C(4, 2) X C(2, 1) X C(1,1) different ways. Therefore, n(B) = C(2, 2) X C(5, 2) X C(4, 2) X C(2, 1) X C(1,1) = C(5, 2) X 4!/2! = 120

Therefore, the total number of four-letter words: n(A) + n(B) = 480
 
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This is the correct answer as I see it .
 
arunbg: that's what i thought also, but my teacher said that it's wrong.. Any idea why?
 
what's ur teacher's answer ?
 
arunbg: My teacher said that it is wrong but she did not tell what is her answer. So i sent her an e-mail asking what is the correct answer, but she hasn't got back to me. And unfortunately, e-mail is the only available form of communication since this course is an online course.

Can I ask anyone else's opinion for this question?

Thank you...
 
ok, I think you should try this:
C=n! / r! (n-r)!
where n is the nuber of symbols, taken r at the time.
therefore, in this case, n=7 and r=4
C= 7! / 4!3! and that's how you get 35 different combinations.
 
Tinaaa said:
ok, I think you should try this:
C=n! / r! (n-r)!
where n is the nuber of symbols, taken r at the time.
therefore, in this case, n=7 and r=4
C= 7! / 4!3! and that's how you get 35 different combinations.

Welcome to PF Tinaaa.
Don't want to start on a sour note, but you might want to reread the question .
 
I've just reread it and I still think that there is nothing wrong with my answer. 35 4-letter words can be formed from the word PULLEYS =) Why do u think it's wrong??
 
Well, firstly you haven't taken into account the no. of ways in which you can arrange the letters to form different words.
Also, you have not taken into account the repitition of the two L's in
PULLEYS which of course cannot be interchanged to get new words.
7C4 only gives you the no of ways of choosing 4 different objects from 7.

I suggest you read the OP's detailed solution to see how it is done.

Regards
Arun
 

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