Permutations?

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JulianMathHelp

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Let's say a letter is LOLIPOP, and it asks how many times you can arrange the word. Is this a permutation problem even though it has repeated countings?
 
JulianMathHelp said:
Let's say a letter is LOLIPOP, and it asks how many times you can arrange the word. Is this a permutation problem even though it has repeated countings?
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The name \(\bf MISSISSIPPI\) has eleven letters, four \(S's\), four \(I's\) & two \(P's\).
That word can be rearranged in \(\dfrac{11!}{(4!)(4!)(2!)}\) ways.
Now you apply that to \(\bf LOLIPOP\) and post the result.
 
I k
pka said:
The name \(\bf MISSISSIPPI\) has eleven letters, four \(S's\), four \(I's\) & two \(P's\).
That word can be rearranged in \(\dfrac{11!}{(4!)(4!)(2!)}\) ways.
Now you apply that to \(\bf LOLIPOP\) and post the result.
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8!/2!2! Why are the denominators factorial instead of just (4)(2)(2)? Is this a combination problem?
 
JulianMathHelp said:
8!/2!2! Why are the denominators factorial instead of just (4)(2)(2)? Is this a combination problem?
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You cannot count. WHY? LOLIPOP has seven letters not eight.
There are two L's, two P's & two O's.
If we subscripted the repeaters \(L_1O_1L_2IP_1O_2P_2\) now there are seven distinct letters.
Those seven distinct letters can be arranged is \(7!\) ways.
If we drop the subscripts then we have a lot of duplicates is those \(7!\) so what must we divide by?
 
JulianMathHelp said:
Divide by 2!2!2!? Why though?
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There's a fairly nice explanation here.

(Note, however that the word is usually spelled LOLLIPOP, which changes things.)

Or you could search for "permutations of multisets", which is what this is -- a set of non-distinct items. (But ignore anything you find about k-permutations.)

But pka was presumably hoping you would think for yourself. Can you come up with a reason before looking it up?
 
Ok, so this is before I read it. When we divide by one 2!, aren't we taking out all of the double counted letters as a whole?
 
Ok, I think I understand it now. Let's say you have the letters "LOLLIPOP". When you divide by 2! (the number of ways to arrange 2 P's), you convert two instances where the "P"'s are swapped, and convert it into one. However, there are still other "multiple" counting of letters within those newly converted permutations. So, we have to divide by 3! as that will take all instances where the 3 "L's" are swapped around without any of the other letters changing, and convert it into 1. Then, this will result in the final answer.
 
JulianMathHelp said:
Divide by 2!2!2!? Why though?
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Let's look at the word \(LOONROOM\) it has eight letters with four repeating.
Subscript the O's, \(LO_1O_2NRO_3O_4M\). Now we have a word with eight distinct letters.
There are \(8!=40302\) ways to rearrange that one eight letter word.
\(O_2 O_1 O_4 O_3MRLN\) is just one having all the \(0's\) at the start.
Here three more: \(O_1 O_2 O_4 O_3MRLN\) \(O_4 O_3 O_2 O_1MRLN\) & \(O_4 O_1 O_2 O_3MRLN\)
That is just four of the \(4!=24\) ways to rearrange the word \(LOONROOM\) having all subscripted \(O's\) at the beginning.
So of the \(40302\) rearrangements of \(LO_1O_2NRO_3O_4M\) by dropping the subscripts we get \(4!\) identical copies.
So divide.
 
pka said:
Let's look at the word \(LOONROOM\) it has eight letters with four repeating.
Subscript the O's, \(LO_1O_2NRO_3O_4M\). Now we have a word with eight distinct letters.
There are \(8!=40302\) ways to rearrange that one eight letter word.
\(O_2 O_1 O_4 O_3MRLN\) is just one having all the \(0's\) at the start.
Here three more: \(O_1 O_2 O_4 O_3MRLN\) \(O_4 O_3 O_2 O_1MRLN\) & \(O_4 O_1 O_2 O_3MRLN\)
That is just four of the \(4!=24\) ways to rearrange the word \(LOONROOM\) having all subscripted \(O's\) at the beginning.
So of the \(40302\) rearrangements of \(LO_1O_2NRO_3O_4M\) by dropping the subscripts we get \(4!\) identical copies.
So divide.
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I perfectly understand instances where only one letter is repeated. I was having an issue when multipel distinct letters were being repeated. I wrote my understanding, could you check it please?
 
JulianMathHelp said:
perfectly understand instances where only one letter is repeated. I was having an issue when multipel distinct letters were being repeated. I wrote my understanding, could you check it please?
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The logic is exactly the same.
How many ways can the string \($$@@@@|||||0000001111111\) be rearranged?
Answer:\(\dfrac{24!}{2!\cdot 4!\cdot 5! \cdot 6!\cdot 7! }\)
 
So if I received the question:
How many ways can "STARWARS" be arranged, it would be 8!/(2!2!) The first 2! is to get rid of the duplicates of "S", and the second is to get rid of the duplicates of "R"'s.
 
JulianMathHelp said:
So if I received the question: How many ways can "STARWARS" be arranged, it would be 8!/(2!2!) The first 2! is to get rid of the duplicates of "S", and the second is to get rid of the duplicates of "R"'s.
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By George, I think you have it.
 
JulianMathHelp said:
So if I received the question:
How many ways can "STARWARS" be arranged, it would be 8!/(2!2!) The first 2! is to get rid of the duplicates of "S", and the second is to get rid of the duplicates of "R"'s.
Click to expand...
How about the duplicate "A"s?
 
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