Number of rearrangements of letters

  • MHB
  • Thread starter mathmari
  • Start date
In summary, the author has calculated the rearrangements of the letters 'WINPRESENT' that contain either the word 'WIN' or the word 'PRESENT' or both of them. They have found that the total amount of rearrangements is $\frac{8!}{2}+4!-2$. If we take into account that we have twice the letter N, the total amount of rearrangements is $\frac{9!}{4}+5!-2\cdot 2$.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
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)
 
  • #4
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)
 
  • #5
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)
 
  • #6
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 $\frac{9!}{4}+5!-2$.
 
  • #7
Ahh yes (Emo)

Thank you! (Sun)
 

FAQ: Number of rearrangements of letters

How do you calculate the number of rearrangements of letters?

The number of rearrangements of letters is calculated using the formula n! / (n1! * n2! * ... * nk!), where n is the total number of letters and n1, n2, ..., nk are the number of occurrences of each letter. This formula takes into account repeated letters in the word.

What is the difference between permutations and combinations?

Permutations refer to the number of ways to arrange a set of objects in a specific order, while combinations refer to the number of ways to choose a subset of objects from a larger set, without considering the order. In the context of rearranging letters, permutations would be used to calculate the total number of possible arrangements, while combinations would be used to calculate the number of unique combinations of letters.

How does the length of the word affect the number of rearrangements?

The length of the word directly affects the number of rearrangements, as the number of possible arrangements increases as the word gets longer. For example, a 5-letter word will have more possible arrangements than a 3-letter word. The formula for calculating the number of rearrangements takes into account the length of the word.

Can the number of rearrangements be larger than the total number of letters in the word?

No, the number of rearrangements cannot be larger than the total number of letters in the word. This is because the formula for calculating the number of rearrangements takes into account the number of repetitions of each letter, and therefore, the maximum number of rearrangements will always be equal to the number of letters in the word.

How can the number of rearrangements be used in real-life applications?

The number of rearrangements of letters can be used in various real-life applications such as creating unique passwords, generating random codes or serial numbers, and in cryptography. It can also be used in genetics to calculate the number of possible genetic combinations in a DNA sequence. Additionally, it can be used in statistical analysis and probability calculations in fields such as economics and finance.

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