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This is related to a mathematical question I was discussing in math.stackexchange.com.

In how many words the letter of word 'AAAEEEBBBDDDCCCC' be arranged so that only 2 vowels always remain together?

How can I properly word this question?

I am asking this because this question is open to more than one interpretation.

One can interpret this question that there can be only one set of two consecutive vowels and all other vowels are isolated. This is what actually meant by the question. [such as CVVCVCV...]

One can interpret this question that there can be (one set of two consecutive vowels) and (another set of two consecutive vowels) [such as VVCVV...]

Another doubt that comes for this question is whether 3 or more consecutive letters are counted or not (such as CVVVCVCV...

How to properly word this question so that the readers must be clear that it is specifically asked with the first interpretation I mentioned?

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  • The word you may use is 'adjacent'. Mathematicians are quite used to it.
    – Keep these mind
    Nov 20 '16 at 20:16
  • thanks. but, how to avoid all these kind of interpretations and specifically target the first interpretation?
    – Kiran
    Nov 20 '16 at 20:17
  • "In how many words [can] the letter[s] of [the] word 'AAAEEEBBBDDDCCCC' be arranged so that only 2 vowel letters are adjacent?" Or 'next to each other' in stead of 'adjacent'.
    – Keep these mind
    Nov 20 '16 at 20:20
  • @Keepthesemind, but this also mean, words like CCVVCVV also right? (where C is used for consonant and V for vowel)
    – Kiran
    Nov 20 '16 at 20:25
  • 2
    OK. "In how many words [can] the letter[s] of [the] word 'AAAEEEBBBDDDCCCC' be arranged with (only) one adjacent pair of vowel letters?"
    – Keep these mind
    Nov 20 '16 at 20:32
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Several problems are preventing clarity:

  • AAAEEEBBBDDDCCCC isn't a word in the ordinary sense, and no arrangement of these letters is a word.
  • Arrange is ambiguous. Is BEED an arrangement?
  • Only and always are inapt here.

Try this:

How many permutations of the string AAAEEEBBBDDDCCCC have exactly two adjacent vowels?

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  • 1
    Precisely. Less is more. Nov 21 '16 at 4:00
  • but, if we word the question as "How many permutations of the string AAAEEEBBBDDDCCCC have exactly two adjacent vowels?", it means, two adjacent vowels repeating 2 times or more [such as VVCVV...] are also valid ., right?
    – Kiran
    Nov 21 '16 at 23:11
  • @Kiran VVCVV has four adjacent vowels.
    – deadrat
    Nov 22 '16 at 3:09
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"How many words can be formed from "AAAEEEBBBDDDCCCC" such that each has exactly one pair of adjacent vowels?" If you are considering any subset as a word, without regard to the dictionary, then this is a problem of multiset permutation.

This might be stated as "how many r-permutations of M are there for M of length n, r from 2 to n, where M is "AAAEEEBBBDDDCCCC". There are compact formal notations for this in combinatorics.

see here for examples

http://homepage.divms.uiowa.edu/~sokratov/2008m150/multiperm.pdf

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Edit [
! I was (and still am) not up to speed on permutation and combination; I just wanted to flag the terms.
The original “permutation” illustration (below) is fine (except that I found an error in it). The “combination” illustration (below) was quite wrong, for the mathematical term; I have changed it.
In everyday English, we use “combination” for something like choosing a car — 3 engine options, 5 paint options and 2 body styles. That is {a meaning of the term “combination”} different from the mathematical one. (It is straightforward; 3*5*2=30.)
In maths, “permutation” and “combination” are particularly about choosing r options from among n. (Choosing the same option twice is excluded. Having repeats among the options is a special case; I am ignoring that here.)
• Permutation
We are choosing 2 options from among 5. Perhaps we are painting a model; we choose one colour for the body, and a second one for the trim.
The formula is nPr = n! / (n-r)!. The base concept is choosing 5 from among 5 — 5*4*3*2*1 = 5! [“5 factorial”]. That is the numerator. If we are choosing only 2, then we chop off the remaining options — …*3*2*1 — which we do by dividing by (n-r)!.
• Combination
One example here is choosing people for a team — 5 options, 3 choices. Again, the basis is 5*4*3, with …*2*1 removed by dividing by (n-r)!, as above. However, this gives us repeated selections of the same people; ABC,ACB,BAC,BCA,CAB,CBA are all the same option. For each such set, we have to reduce it to 1 option. We do this by [also] dividing by r! (here 3*2*1).
nCr = n! / (n-r!) / r!
]


[As ever, I really hate the fact that “the internet” insists that I could not possibly ever, ever, ever want more than one space.]

You want to know the terms “permutation” and “combination”.

Permutation is like this. For {ABCD} (n=4,r=4):
ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD …

Combination is like this. For {ABCD} (n=4,r=2):
AB AC AD BA BC BD CA CB CD DA DB DC.

These are mathematical terms, so for “permutation”… for (for instance) “AAAB”… with n=4, r=3, “AAB” would appear 6 times. [This [“6”] is offhand; I am not addressing the maths for duplicated options here.]

If you want to prohibit duplicates (e.g. AAB and AAB), you add the word “unique”, which is a strong word that means that, for a given thing, there is only one in the universe (of consideration). [I personally allow “very unique”, meaning not only unique but also strikingly different from everything else, but some people think that “very unique” is a contradiction.] …Or you can just say (add) that duplicates are prohibited.

“Adjacent” means “next to each other”.

Your formulation (“In how many words… always remain together?”) is actually pretty good. The difficulty is that, although it explains what you want, it is not strongly clear and precise, so the reader is a bit nervous about it. You want to use standard mathematical jargon here. (Do a search for puzzles with “permutation” or “combination”.)

[I (personally) am very strict with such things, so maybe the following is not representative [most people].]
The other issue is “word”. I [personally] would take that to mean an actual English word, even in a mathematics puzzle… although I would immediately think it odd. Indeed, I am inferring that here you do not mean “word” in that sense. [Actually, I think people do use “word” in this context meaning “any combination of letters”, but I certainly would not, myself. (Or I might use “ “word” ” (i.e. in scare quotes), if I was being lazy.)]

However, that problem is automatically solved if you use “permutation” or “combination”. (Failing that, I would say, “strings of letters”.)

So… “
How many permutations of the following letters — XXXX — are there… with the following constraint? [Generally: “The constraints are as follows.”]
• Vowels must always stand alone — i.e. be surrounded by consonants — except for one instance that can have 2 adjacent vowels.

It is more wordy, but that gives the reader confidence.

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