How many 7-letter words can I form, starting with X, containing one and only one vowel, and considering letters can be repeated?

How do I solve this using counting and/or permutations?

I saw an answer for this but there were $ signs and I was not able to understand it.

  • 1
    $\begingroup$ Solve what? You haven't asked a question. Do you want a count of such words? Is it exactly one vowel, or possibly more? $\endgroup$ – Thomas Andrews Oct 3 '15 at 14:48
  • $\begingroup$ I apologize for not being clear enough! How many 7-letter words can I form based on those criteria? And only one vowel $\endgroup$ – user276520 Oct 3 '15 at 14:53

The $X$ at the beginning is fixed, and letters can be repeated, so we can just ignore it.

The vowel can be in one of $6$ places and there are $5$ possible vowels. The other $5$ letters can each be one of the $21$ consonants.

So in total there are $6\cdot 5 \cdot 21^5=122523030$ such words.


If the letters can't be repeated, consider the $5$ remaining letters:

The first one can be one of $20$ letters (not an $X$, not a vowel). The second one can be one of $19$ letters (same as first, but can't equal the first either). etc.

So there are $6\cdot 5\cdot 20\cdot19\cdot 18\cdot 17\cdot 16=55814400$ words.

  • $\begingroup$ What if letters cannot be repeated? $\endgroup$ – user276520 Oct 3 '15 at 14:59
  • $\begingroup$ @user276520 I edited my answer for the non-repeated case. $\endgroup$ – Alex Oct 3 '15 at 15:03
  • $\begingroup$ Everything's clear now. Thank you very much Alex! $\endgroup$ – user276520 Oct 3 '15 at 15:05
  • $\begingroup$ @user276520 You're welcome :) $\endgroup$ – Alex Oct 3 '15 at 15:06

The basic idea here is the "fundamental theorem of counting": if event A can happen in m ways and, for each of those, event B can happen in n ways, then both events can happen in mn ways.

Since the first letter must be X, it is just the 6 succeeding letters that can be changed. There are 5 vowels so, assuming we put the vowel as the second letter, there are 5 choices. There are 21 "non-vowels" so there 21 choices for the last 5. The number of possibilities for X, vowel, and then 5 non-vowels is 5(21)(21)(21)(21)(21)= 5(21^5). But we could have the same letters with the vowel in any of the six places following X so we must multiply that by 6: 5(6)(21^5).

(This is assuming that "contains one vowel" means "contains exactly one vowel".)


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