Combinatorics - Permutations of abcdefg. Which is right?

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Homework Help Overview

The discussion revolves around determining the number of permutations of the letters a, b, c, d, e, f, g with specific conditions regarding the placement of letters a and b, specifically having either two or three letters between them.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore different cases based on the number of letters between a and b, with one participant attempting to calculate permutations for various scenarios. Others question the validity of considering fewer than seven letters and clarify the focus on permutations rather than combinations.

Discussion Status

The discussion includes attempts to clarify the problem's requirements and the correct interpretation of permutations. Some participants express uncertainty about the original poster's calculations and the definitions being used, while others provide feedback on the approach taken.

Contextual Notes

There is a noted confusion regarding the distinction between permutations and combinations, with participants reflecting on the problem's phrasing and their understanding of the requirements.

Alche
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Homework Statement


How many permutations of the letters a, b, c, d, e, f, g have either two or three letters between a and b? b _ _ a is also very much possible.

Homework Equations


nPr= n!/(n - r)!, where n >= r

The Attempt at a Solution



For this question there can be 4 cases which are as follows

1)when there are 4 letter words,

a _ _ b

from among 5 remaining letters 2 can be taken in 5P2 ways which can be arranged themselves in 2! ways and a and b can also be arranged among themselves in 2 ways, so

5P2*2!*2! = 80

2)5 letter words, here can be 3 cases too which are as follow:-

A) a _ _ b _
B) a _ _ _ b
c) _ a _ _ b

letters can be arranged here as

(5P3*3!*2)*3 = 2160.

3) when 6 lettered words are formed

a) a _ _ _ b _
b)a _ _ b _ _
c)_ a _ _ b _
d)_ a _ _ _ b
e) _ _ a _ _ b

here the letters can be arranged as

(5P4*4!*2)*5 = 28800

4)when 7 lettered word is formed

a) a _ _ _ b _ _
b) a _ _ b _ _ _
c) _ _ a _ _ _ b
d) _ _ _ a _ _ b
e) _ a _ _ b _ _
f) _ a _ _ _ b _
g) _ _ a _ _ b _

(5P5*5!*2)*7 = 201600

So now am getting the answer as 232640.

Please tell me am I right? If not then where am I making mistake.
 
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I think you are wrong as there are no 4 letter words. All your words are 7 letters (we are talking about permutations after all). The question is: how many do we have with your restriction. When a and b are in the following place

a _ _ b _ _ _

we have 5! words (as there are 5 places left for the remaining letters). However a and b can be placed in more ways. Count them too. Your answer at 4) is in the good direction.

my answer
1680
 


Thanks for replying.

I got it.
 
Last edited:


Apparently Alche is NOT talking about "permutations" but about both permutations and combinatiions combinations of 4 to 7 letters, chosen from a, b, c, d, e, f, g, with "either two or three letters between a and b".
 


HallsofIvy said:
Apparently Alche is NOT talking about "permutations" but about both permutations and combinatiions combinations of 4 to 7 letters, chosen from a, b, c, d, e, f, g, with "either two or three letters between a and b".

But in the question it is only mentioned about permutations of the letters, they have no where asked about the combination. I think I was not able to grasp what it was asking earlier, now I got it.

If am wrong then please correct me.
 

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