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The total number of words are: $5^4.$
Am unclear as how to find the words of size $4,$ with one vowel, out of the two available.
I could only muster the calculation of: $3*3*3*2 = 54.$ While have got a compilation of such 'actual' words, which while still being incomplete, has crossed the count of $64.$

So, request the logic for correct computation.

jiten
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2 Answers2

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There are $3 \cdot 3 \cdot 3 \cdot 2 = 54$ words with one vowel in the last position. The vowel can be in any of the four positions giving $54 \cdot 4 = 216$ possibilities.

Steven
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  • Request to also solve it using Inclusion-exclusion method. – jiten Jan 03 '25 at 14:28
  • @jiten why would you want that? I think that would make the argument more complex, this is about as simple as it gets. – Steven Jan 03 '25 at 14:34
  • For learning reasons only, wanted a universal approach, like the IE method. – jiten Jan 03 '25 at 14:36
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    Well this can be seen as a trivial case of inclusion-exclusion. Let $E_i$ denote the set of four letter words that have one vowel in the $i$th position. Then the number of four letter words with one vowel is $|E_1| + |E_2| + |E_3| + |E_4| - |E_1 \cap E_2| - |E_1 \cap E_3| - \cdots$, but all those intersections are empty. So this is more or less too simple to even apply inclusion-exclusion. – Steven Jan 03 '25 at 14:45
  • Thanks, and request to tell how to modify your last comment for the above problem, for the two vowels' case. I mean that cannot adapt your last comment for that case, as $E_i$ cannot be used in that case. – jiten Jan 04 '25 at 07:50
  • @jiten you could use $E_{i,j}$ for $i\neq j$, which you define as the four letter words with vowels in positions $i$ and $j$. – Steven Jan 04 '25 at 10:56
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The original poster left a comment requesting an Inclusion-Exclusion solution. This is given below. For what it's worth, I regard Inclusion-Exclusion to be an inferior method of attack, for this particular problem.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Following the syntax in the 2nd link above, let $~S~$ denote the set of all possible words, that contain no more than $~2~$ vowels.

Let $~S_a~$ denote the subset of $~S~$ that contains exactly one "a".
Let $~S_e~$ denote the subset of $~S~$ that contains exactly one "e".

Then, you want

$$| ~S_a \cup S_e ~| - | ~S_a \cap S_e ~|. \tag1 $$

Further, by inclusion exclusion theory,

$$| ~S_a \cup S_e ~| = | ~S_a ~| + | ~S_e ~| - | ~S_a \cap S_e ~|. \tag2 $$

Putting (1) and (2) above together, the desired computation is

$$| ~S_a ~| + | ~S_e ~| - 2| ~S_a \cap S_e ~|. \tag3 $$

Note
More usual in similar Inclusion-Exclusion problems is to define $~S~$ to equal the set of all words, without any regard for how many vowels there are. Unfortunately, with the more usual definition of $~S,~$ you would have (for example) the subset $~S_a~$ including words with exactly $~1~$ "a", but more than $~1~$ "e".

In such an event, the formula shown in (1) above would not represent the desired computation, so the formula shown in (3) above would also not represent the desired computation (i.e. the answer to the problem). So, to make the inferior Inclusion-Exclusion approach work, I resorted to the somewhat artificial definition of $~S.$

By symmetry,

$$| ~S_a ~| = | ~S_b ~| $$

$$= \binom{4}{1} \times \left\{ ~\left[ ~\binom{3}{1} \times 3^2 ~\right] + \left[ ~\binom{3}{0} \times 3^3 ~\right] ~\right\} = 216.$$

That is, there are $~4~$ choices for where to put the "a" and then two mutually exclusive possibilities:

  • $~3~$ choices for where to put the "e", and then $~3^2~$ ways of completing the word.

  • No "e" is used, so there are $~3^3~$ ways of completing the word.

Also,

$$| ~S_a \cap S_b ~| = \binom{4}{1} \times \binom{3}{1} \times 3^2 = 108.$$

That is, there are $~4~$ choices for where to put the "a", then $~3~$ choices for where to put the "e", and then $~3^2~$ ways of completing the word.

So, the final computation is

$$2 \times [ ~216 - 108 ~] = 216.$$

user2661923
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  • Please elaborate how the computation of $|S_a\cap S_b|$ is done. – jiten Jan 04 '25 at 07:54
  • @jiten Since $~S_a \cap S_b~$ represents the subset of $~S~$ where there is exactly one "a", and exactly one "e", you have $~4~$ choices for which position in the word that the "a" will occur. Once the "a" has been positioned, there will then be $~3~$ remaining positions that the "e" can occur. Then, once the "a" and "e" have been positioned, each of the two remainining positions must be populated by one of the three consonants. – user2661923 Jan 04 '25 at 14:32