Solving Combination Question: Word Length & Alphabetical Order

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Discussion Overview

The discussion revolves around a combination problem involving the formation of words of length 4 from letters in increasing alphabetical order. Participants explore the reasoning behind a provided solution and discuss various approaches to counting acceptable combinations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the provided solution and seeks clarification on the calculation's purpose and the cases it represents.
  • Another participant suggests that there are multiple methods to approach the problem, emphasizing the need for an easier counting strategy.
  • A participant proposes breaking down the problem into smaller groups, noting that for any 4 distinct letters, there is only one acceptable ordering.
  • There is a suggestion to consider the counting of strings with repetitions and how they fit into the overall solution.
  • One participant critiques another's proposed formula, indicating it would lead to a complex sum that is not straightforward to relate to the original answer.
  • Another participant reflects on starting over with a simpler case to understand the combinations better.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to solve the problem, with multiple competing views and methods being discussed. There is ongoing uncertainty regarding the clarity and correctness of the proposed solutions.

Contextual Notes

Participants express varying levels of familiarity with permutations and combinations, which may affect their understanding of the problem and the proposed solutions. The discussion includes attempts to clarify the reasoning behind specific terms in the solution without resolving the overall approach.

Batmaniac
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I have a fully solved combination question that I can't seem to follow.

"How many words of length 4 can be formed if the letters in the word must be in increasing alphabetical order? Note: {A,A,D,E} is but {A,D,A,E} is not"

The solution is as follows (C denotes choose)

(26 C 1) + (26 C 1)(25 C 1) + (26 C 2) + (26 C 1)(25 C 2) + (26 C 4) = 23751

I am having trouble seeing just what this calculation is doing and why it solves the problem (I've only just recently been introduced to permutations and combinations).

To my understanding, each term in the solution is its own case, but I'm not sure what these cases are.

Any help in explaining this solution would be appreciated.

Thanks.
 
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Can you post the thought process that led to this solution?
 
There are probably several ways to do this. You really need to find some way of making the counting easy.

You could set up all kinds of difference equations to solve this, but that is hard. You could sum over all possible starting letters, but that is hard.

How about splitting things up into smaller groups. It is easy to calculate the number of acceptable strings with no repetitions: given any 4 distinct letters, there is precisely one acceptable ordering. Or if you like, there are 26 choose 4 acceptable strings. Now, what about the ones we've yet to count?
 
EnumaElish said:
Can you post the thought process that led to this solution?

It isn't a solution they devised, so probably not. But it is breaking it down into cases - the first term counts the strings with one letter in them, and so on.
 
Oh, I hadn't seen the "a" in "I have a fully solved". Sorry.

I am guessing that the answer is:

letter1 * (26 - rank(letter1) + 1) * (26 - rank(letter2) + 1) * (26 - rank(letter3) + 1) summed over letter3, then over letter2, then over letter1.
 
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Erm, that is not the way to do it. It would contain approximately a thousand or so terms in the sum, wouldn't it? I mean, it is certainly large, and certainly not obvious how to relate it to the given answer which is easy to explain (follow my hint above).
 
I thought my formula would resolve into combinations.
 
If it's correct it will. But can you see an easy way to do that?
 
Point taken. Starting over, with one letter (e.g. AAAA), there are 26 = C(26, 1) combinations.
 
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