Strings of eight English letters

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Discussion Overview

The discussion revolves around the combinatorial problem of determining the number of strings of eight English letters that contain exactly one vowel, with the condition that letters can be repeated. Participants explore different approaches to solving the problem and analyze the implications of the wording in the question.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the number of strings containing at least one 'a' and no other vowels, arriving at a total of $5(22^8-21^8)=85,265,070,875$, but questions the discrepancy with the book's answer of $72,043,541,640$.
  • Another participant proposes a method where they first place the vowel in one of the eight positions, calculating the total as $N=8\cdot(5\cdot21^7)=72043541640$, which matches the book's answer.
  • There is a concern raised about the interpretation of the problem, specifically whether the vowel can occur more than once, given that letters can be repeated.
  • Clarification is sought regarding the interpretation of "exactly one vowel," with one participant asserting that it means only one of the eight letters can be a vowel.

Areas of Agreement / Disagreement

Participants express differing interpretations of the problem, particularly regarding the condition of having "exactly one vowel." There is no consensus on the correct approach or interpretation of the problem, leading to multiple competing views.

Contextual Notes

Participants' calculations depend on their interpretations of the problem statement, particularly the meaning of "exactly one vowel" and the implications of letter repetition. The discussion highlights the potential for misunderstanding in combinatorial problems.

alexmahone
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How many strings of eight English letters are there that contain exactly one vowel, if letters can be repeated?

My attempt:

Let us first find the number of strings that contain at least one 'a' and no other vowels.
Total number of strings including 'a' but excluding the other vowels $=22^8$
Number of strings without any vowels $=21^8$
So, number of strings that contain at least one 'a' and no other vowels $=22^8-21^8$

Sincere there are five vowels, we get $5(22^8-21^8)=85,265,070,875$. The book's answer is $72,043,541,640$. Where did I go wrong?
 
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This is how I would approach the problem...

Suppose the vowel is first, so we have 5 choices for the first letter, and the 21 letters for each of the remaining 7 letters, and applying the counting rule, we find:

$$N_1=5\cdot21^7$$

But now, we consider that the vowel may be anywhere, in 1 of 8 places, hence:

$$N=8\cdot N_1=72043541640$$
 
MarkFL said:
This is how I would approach the problem...

Suppose the vowel is first, so we have 5 choices for the first letter, and the 21 letters for each of the remaining 7 letters, and applying the counting rule, we find:

$$N_1=5\cdot21^7$$

But now, we consider that the vowel may be anywhere, in 1 of 8 places, hence:

$$N=8\cdot N_1=72043541640$$

Thanks.

But can't the vowel occur more than once? After all the question says "if letters can be repeated".
 
Alexmahone said:
Thanks.

But can't the vowel occur more than once? After all the question says "if letters can be repeated".

The problem states:

How many strings of eight English letters are there that contain exactly one vowel, if letters can be repeated?

I took this to mean that only one of the eight letters can be a vowel. :)
 

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