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[SOLVED] Number of rearrangements of letters

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,687
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,687
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,687
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.

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 $\frac{9!}{4}+5!-2$.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
Ahh yes (Emo)

Thank you!! (Sun)