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Saumyojit

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how many 3 letter words can be made using {a,a,b,b} Repetiton of same arrangement not allowed

MY SOLUTION: each unique arrangement of 3 letters word can be made using {a,a,b,b} with no duplicates denoted by x1 up to Xn ={x1,x2...Xn}
for each unqiue arrangement i will get one extra duplicate denoted by the other set {x1...Xn}

SO total permutations with repetions = 4p3 =24 words

24 ={x1,x2..Xn} + {x1,..Xn}
all permutations including reputations=unique arrangments+ each dupilcate of each unique arrangement

24 ={x1,x2..Xn} + {x1,..Xn} ---> aab will produce one duplicte or abb will produce one extra so on
adding=>24=2{x1,x2..Xn}
{x1,x2..Xn}=>24/2=12 words
n=12

BUT ANSWER IS 6.
@Dr.Peterson @HallsofIvy @Jomo
 
How many ways can you choose a letter that will be used twice in the word?

Two, a or b.

Given that one letter is to be used twice in a three-letter word, how many positions can be chosen for the other letter?

Three, first position, second position, or third position.

3 * 2 = 6.
 
JeffM said:
How many ways can you choose a letter that will be used twice in the word?
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WHats wrng with the original question "how many 3 letter words can be made using {a,a,b,b} Repetiton of same arrangement not allowed "

why am I not getting the unique arrangements using this method . This method works for all the sums that i have done of reaarangign words type of problemn sum.
24 ={x1,x2..Xn} + {x1,..Xn}
all permutations including reputations=unique arrangments+ each dupilcate of each unique arrangement



24 ={x1,x2..Xn} + {x1,..Xn} --> for one x1 of the first set we will get another duplicate x1 that is in the other set.
Why 6 is not coming . It will only come if we divide 24 by4
 
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I can't understand your notation, so I can't tell you what is wrong with your reasoning.

Perhaps try to explain it with reference to the actual list of permutations you were shown. What are the two sets?
 
Dr.Peterson said:
I can't understand your notation, so I can't tell you what is wrong with your reasoning.

Perhaps try to explain it with reference to the actual list of permutations you were shown. What are the two sets?
Click to expand...
waiting for u reply. u finally came .thanks

THe first set {x1..Xn} ---denotes all the distinct arrangements aab,bba,aba...so on. THe second set {x1..Xn} or u could take{Y1...Yn} denotes the duplicates of each unique arrangements if we change the position of a1 a2 or b1 b2.
added both of them and equated it to 24.
 
Here is the mistake you are making: a_1 is indistinguishable from a
Saumyojit said:
WHats wrng with the original question "how many 3 letter words can be made using {a,a,b,b} Repetiton of same arrangement not allowed "

why am I not getting the unique arrangements using this method . This method works for all the sums that i have done of reaarangign words type of problemn sum.
24 ={x1,x2..Xn} + {x1,..Xn}
all permutations including reputations=unique arrangments+ each dupilcate of each unique arrangement



24 ={x1,x2..Xn} + {x1,..Xn} --> for one x1 of the first set we will get another duplicate x1 that is in the other set.
Why 6 is not coming . It will only come if we divide 24 by4
Click to expand...
There's nothing wrong with the original question other than it is vaguely worded.

The problem with your reasoning is that you are ignoring the fact that the letters A are indistinguishable as are the letters B. Suppose the letters are differently colored. So we have one red A, one blue A, one red B, and one blue B. We can construct 2 * 2 variants of AAB if we distinguish words by colors in addition to letters.

AAB
AAB
AAB
AAB

That would give you 24 permutations. 4 ways to pick the first colored letter, 3 ways to pick the second colored letter, and 2 ways to pick the third colored letter: 4 * 3 * 2 = 24.

But the letters are not colored. All four of those listed above reduce to 1, namely AAB. That's where the division by 4 comes in.

I really think it is far easier to analyze using the basic counting principles.

The nature of the problem requires that two letters in the three-letter word be the same because, of the four letters available, two are indistinguishable a's, and two are indistinguishable b's. So there are two ways to choose the repeated letter. There are three positions where the non-repeated letter can be slotted. 3 * 2 = 6. I do not say that is the only way to analyze the problem correctly, but it sure is the easiest.
 
If you think the answer is 24, or any other number, then you should be able to list those 24 words. If you can't then your answer is wrong. Did you try to list the possible words? How many were you able to list?
 
JeffM said:
There's nothing wrong with the original question other than it is vaguely worded.
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THere is nothing wrong with the question . It was my mistake that i could not think properly.its easy actually. {a1,a2,b1,b2}

for each unqiue arrangement i will get 3 extra duplicate . a1a2b1 ,a1a2b2,a2a1b1,a2a1b2 .
Ok its solved.
 
There are 2 cases:
1: Permutation where repetitions are not allowed means we cannot count the" extra repetitions of same arrangements"
2: Permutation where repetitions are allowed means we can use one of the n things to be filled r times . eg: making a 2 letter word using a,b where n is {a,b} gives =aa,ab,ba,bb -->in aa and bb we have repeated one of the n things r times(2 times) in two arrangements out of 4 .

CONCLUSION: in the first case repetitions means "arrangement" and second case means "repetion of one of the n things r times(2 times) in some arrangements
is my assessment correct? @JeffM @Dr.Peterson
 
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I don't know. I told you that I believed the question as given was not well worded.

I tend to think about "distinct instances" rather than "repetitions allowed" and "repetitions not allowed"

We have two a's available. But they are not distinguishable. Thus, I think it confuses matters to talk about a1 and a2.

But there are many ways to skin a cat. You should think about it in any way that makes sense to you.
 
JeffM said:
We have two a's available. But they are not distinguishable. Thus, I think it confuses matters to talk about a1 and a2.
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@JeffM
Have u understood what was i trying to refer to by the word "repetition" denoting different things in case 1 and case 2 in post no10
CASE 1: find the Permutation of word lle where repetitions are not allowed means it is telling us to count only distinct arrangements lle, ell,lel excluding the other 3 duplicates.

WHERE AS Permutation where repetion are allowed means we can use one of the n things to be filled r times . eg: making a 2 letter word using a,b where n is {a,b} gives =aa,ab,ba,bb -->in aa and bb we have repeated one of the n things r times(2 times) in each of two arrangements out of total 4 . ---> So here repetiton does not mean it is denoting the duplicate arrangements but the repetititon of letters in some arrangements.

Thats the difference i am trying to state!
 
The real distinction that matters is whether the "repetitions" (duplicates) are in the source (your aabb), which is properly called a multiset, or in the result when the source is a set (no duplicates) but the result may have more than one of an item. If no repetition is allowed in the result, it is called a permutation (or sequence) if order matters, or a combination (or set) if order doesn't matter). When the source is a multiset, as in your example, you would never forbid "repetition" in the result, as there would be no meaning in the multiplicity of elements.

I'm not sure the terminology with regard to multisets or "repetition" is entirely standardized, as I've always had trouble finding good sources for information about odd cases like what you are asking about. One term for your "aab" is a 3-permutation of the multiset {a, a, b, b}. You'll observe that no formula is given for this case, only for a permutation using all elements (an anagram).

When the source is a set and repetition is allowed, that should not be called a permutation with repetition, but an n-tuple or a "word"; but as you can see, the term is used.
 
@Dr.Peterson
In the original quest of the post ->making 3 letter word using {a,a,b,b} the multiset has duplicates and we are doing 3 permutation of the mutliset making sure there is one unique arrangement. I know this question a little part falls under permutation with repetion as there are duplicates in the multiset but in finding the answer the process we do is of permutation of multisets .


"When the source is a set and repetition is allowed we can call it anyone name whether tuple or permutation with repetition" .
i am not asking about what is permutation with repetition and permutation without repetion!!!. I know how the process works and when it works. I also know the difference between p& C. Please try to understand once again

I was saying that i discovered the subtle difference in the meaning of the word " repetiton" in two cases

1st case: (Permutation of multisets or permutation without repetion)--> Repetiton here means not allowing duplicates of each unique arrangement to be counted. okay?

2nd case: Permutations with repetition or n-tuples --> Repetiton here means we can "repeat" one of the n things to be filled r times . eg given in post #12
 
Please don't shout. If you're trying to tell me what English means, my answer is, already, that English is ambiguous. Perhaps we agree on that.

But we never count "duplicates of the same arrangement", so I think it is not helpful to use the term at all in that way. Can you quote a reputable source that uses the word as you are using it in case 1, or are you only talking about your own personal use of the word?
 
I am sorry if u got hurt by exclamanation marks. I am talking about what i have understood with the word repetition in both the two cases . noticed a Sublte differnece.
In the first case(Perm. without repetition) means we cannot count or never count duplicates of the arrangments .Yeah thats what permutation means

In the second case (Permutations with repetition or n-tuples) --> Repetiton here means we can "repeat" one of the n things to be filled max r times in some arrangements of the whole permutations . eg given in post #12
 
The real distinction that matters is whether the "repetitions" (duplicates) are in the result when the source is a set (no duplicates) but the result may have more than one of an item.
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@JeffM @Dr.Peterson by this line u mean if i assume a eg of a set which contains no duplcates {a,b}-->question is how many 2 letter words using this set can we make repetion allowed-->result is {aa,bb,ab,ba}-->so u are saying that "if the set does not contain any duplicates but the result may have more than one of an item" .In the result these two arrangements {aa,bb} proves ur point. Is my assumption right?

When the source is a multiset, as in your example, you would never forbid "repetition" in the result, as there would be no meaning in the multiplicity of elements.
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using the same question(how many 2 letter words using this multiset can we make) --> Now The source is a multiset in this case {a,a,b,b} 2 letter permutations will come-->{aa,ba,ab,bb } 2 out of 4 arrangments in the result contains one repetiton of one letter--> This is what u are trying to say by "you would never forbid "repetition" in the result" right?
 
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@Dr.Peterson please also reply to my post 17 that i posted 20 hours ago.I just told the difference in the meaning of word "repeation" of what i observed. thats it
 
Saumyojit said:
I was saying that i discovered the subtle difference in the meaning of the word " repetiton" in two cases

1st case: (Permutation of multisets or permutation without repetion)--> Repetiton here means not allowing duplicates of each unique arrangement to be counted. okay?

2nd case: Permutations with repetition or n-tuples --> Repetiton here means we can "repeat" one of the n things to be filled r times . eg given in post #12
Click to expand...

My point was that we simply don't use the term "without repetition" to mean "not allowing duplicates of an arrangement". Counting means counting unique arrangements, so there is no need to say whether duplicate arrangements are allowed. "Without repetition" always refers to the elements within an arrangement.

As far as I can tell, what you "discovered" is not about actual usage of the word, but about your own wrong usage.

But, frankly, I am just tired of the discussion; and if I choose not to answer you, that is my privilege.
 
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