Number of rearrangements of letters

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Discussion Overview

The discussion revolves around calculating the number of rearrangements of the letters in 'WINPRESENT' and 'WINPRESENTS' that contain the subwords 'WIN' or 'PRESENT' or both. The scope includes combinatorial reasoning and mathematical calculations related to permutations of letters.

Discussion Character

  • Mathematical reasoning, Exploratory, Debate/contested

Main Points Raised

  • One participant calculates the rearrangements containing 'WIN' as $8!$ and those containing 'PRESENT' as $\frac{8!}{2}$, leading to a proposed total of $\frac{8!}{2}+4!-2$.
  • Another participant agrees that for the initial problem, the double occurrence of 'N' does not need to be accounted for.
  • A later post introduces 'WINPRESENTS' and suggests a new calculation involving $9!$ and $\frac{9!}{4}$ for 'PRESENTS', questioning if the total should be $\frac{9!}{4}+4!-2$.
  • There is a suggestion that the total should account for double counting, leading to a proposal of $-2\cdot 2$ for the overlaps.
  • One participant corrects themselves regarding the subtraction of overlaps, stating it should be $-2$ instead of $-2\cdot 2$.
  • Another participant raises a new example with 'KERDISESDWRO', calculating rearrangements for 'KERDISES' and 'DWRO', and questioning the correctness of their calculations.
  • There is a clarification about the correct terms to use in the calculations, ensuring consistency in the subwords referenced.

Areas of Agreement / Disagreement

Participants generally agree on the approach to calculating rearrangements but express uncertainty about specific details and corrections. Multiple competing views and calculations remain present throughout the discussion.

Contextual Notes

Participants express uncertainty regarding the treatment of repeated letters and the correct application of combinatorial principles in their calculations. The discussion includes evolving calculations and corrections based on earlier posts.

Who May Find This Useful

Readers interested in combinatorial mathematics, particularly those exploring permutations and arrangements of letters, may find this discussion relevant.

mathmari
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Hey! :o

We have the letters 'WINPRESENT'. I want to calculate the rearrangements of these letters that contain either the word 'WIN' or the word 'PRESENT' or both of them.

I have done the following:

The subword 'WIN' is contained in $8!$ rearrangements. Since at 'PRESENT' we have twice the letter E we get $\frac{8!}{2}$ rearrangements. The subword 'PRESENT' is contained in $4!$ rearrangements. We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'WIN' or the word 'PRESENT' or both of them is equal to $\frac{8!}{2}+4!-2$, or not?

Or do we have to take into account that we have twice the letter N ?

(Wondering)
 
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mathmari said:
Hey! :o

We have the letters 'WINPRESENT'. I want to calculate the rearrangements of these letters that contain either the word 'WIN' or the word 'PRESENT' or both of them.

I have done the following:

The subword 'WIN' is contained in $8!$ rearrangements. Since at 'PRESENT' we have twice the letter E we get $\frac{8!}{2}$ rearrangements. The subword 'PRESENT' is contained in $4!$ rearrangements. We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'WIN' or the word 'PRESENT' or both of them is equal to $\frac{8!}{2}+4!-2$, or not?

Or do we have to take into account that we have twice the letter N ?

Hi mathmari!

I believe you are quite correct.
And indeed, for this problem we shouldn't take into account that N occurs twice.
That only becomes relevant if we want to know for instance all possible configurations. Then we'd have to compensate for double-counting N. (Nerd)
 
I like Serena said:
I believe you are quite correct.
And indeed, for this problem we shouldn't take into account that N occurs twice.
That only becomes relevant if we want to know for instance all possible configurations. Then we'd have to compensate for double-counting N. (Nerd)

Ah ok!

If we consider the letters 'WINPRESENTS' (now with an extra S at the end) then we have the following:
The subword 'WIN' is contained in $9!$ rearrangements. Since at 'PRESENT' we have twice the letter E and the letter S do we get $\frac{9!}{4}$ rearrangements?
The subword 'PRESENTS' is contained in $4!$ rearrangements. We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'WIN' or the word 'PRESENT' or both of them is equal to $\frac{9!}{4}+4!-2$, or not?

(Wondering)
 
mathmari said:
Ah ok!

If we consider the letters 'WINPRESENTS' (now with an extra S at the end) then we have the following:
The subword 'WIN' is contained in $9!$ rearrangements. Since at 'PRESENT' we have twice the letter E and the letter S do we get $\frac{9!}{4}$ rearrangements?

The subword 'PRESENTS' is contained in $4!$ rearrangements. We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'WIN' or the word 'PRESENT' or both of them is equal to $\frac{9!}{4}+4!-2$, or not?

(Wondering)

Shouldn't that be $\frac{9!}{4}+4!-2\cdot 2$ (assuming that you meant PRESENTS instead of PRESENT in all cases)? (Wondering)
 
I like Serena said:
Shouldn't that be $\frac{9!}{4}+4!-2\cdot 2$ (assuming that you meant PRESENTS instead of PRESENT in all cases)? (Wondering)

Do we get $2\cdot 2$ because we have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'? (Wondering)

If we consider the letters 'KERDISESDWRO' then we have the following:
The subword 'KERDISES' is contained in 5! rearrangements.
The subword 'DWRO' is contained in 9! rearrangements. Since at 'KERDISES' we have twice the letter E and the letter S do we get $\frac{9!}{4}$ rearrangements? We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'KERDISES' or the word 'DWRO' or both of them is equal to $\frac{9!}{4}+5!-2\cdot 2$.

Is everything correct? (Wondering)
 
mathmari said:
Do we get $2\cdot 2$ because we have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'? (Wondering)

Sorry. (Blush)

It's $-2$ after all.
We have calculated twice 'WINPRESENTS' and twice 'PRESENTSWIN', while we should only count them once.
So we have to subtract 1 for each of them.

mathmari said:
If we consider the letters 'KERDISESDWRO' then we have the following:
The subword 'KERDISES' is contained in 5! rearrangements.
The subword 'DWRO' is contained in 9! rearrangements. Since at 'KERDISES' we have twice the letter E and the letter S do we get $\frac{9!}{4}$ rearrangements? We have calculated twice 'WINPRESENT' and twice 'PRESENTWIN'.

So the total amount of rearrangements that contain either the word 'KERDISES' or the word 'DWRO' or both of them is equal to $\frac{9!}{4}+5!-2\cdot 2$.

Is everything correct?

You're referring to WINPRESENT and PRESENTWIN when that should be KERDISESDWRO and DWROKERDISES. (Nerd)
And we should have $-2$ as before.
So $\frac{9!}{4}+5!-2$.
 
Ahh yes (Emo)

Thank you! (Sun)
 

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