Solving Combination Question: Word Length & Alphabetical Order

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In summary: That I understand. Now, adding a second letter (e.g. AAAB), there are still C(26, 1) combinations, but now they must be divided into two piles, those with A, and those with B. The number of combinations in the A pile is 26 - 1 = 25, and the number in the B pile is 26 - 2 = 24. So the number of combinations with two letters is C(26, 1) * C(25, 1) = C(26, 1) * C(25, 1). I'm struggling to see the connection.In summary, the conversation discusses a fully solved combination question where the
  • #1
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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  • #2
Can you post the thought process that led to this solution?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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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  • #6
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).
 
  • #7
I thought my formula would resolve into combinations.
 
  • #8
If it's correct it will. But can you see an easy way to do that?
 
  • #9
Point taken. Starting over, with one letter (e.g. AAAA), there are 26 = C(26, 1) combinations.
 
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1. How do you solve combination questions involving word length and alphabetical order?

To solve these types of questions, start by organizing the words in alphabetical order. Then, count the number of letters in each word and write it next to the word. Finally, compare the numbers to determine the correct order.

2. Can the words be in different cases (upper or lower) when solving combination questions?

Yes, the words can be in different cases as long as the letters are in the correct order. However, it may be easier to solve the question if all words are in the same case.

3. How many words can be arranged in alphabetical order?

The number of possible combinations for arranging words in alphabetical order is dependent on the number of words and their lengths. Generally, the more words and the longer the words, the higher the number of possible combinations.

4. Are there any shortcuts or strategies for solving these types of questions?

One possible strategy is to first look for any words that begin with a letter that is unique among the other words. This can help narrow down the possibilities and make it easier to determine the correct order.

5. Can these types of questions be solved without writing out all the possible combinations?

Yes, these questions can be solved without writing out all the possible combinations. By using strategies such as the one mentioned above, or by using logic and deductive reasoning, the correct order can be determined without having to write out every possible combination.

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