# 7-Letter word; starts with X; contains one vowel; letters can be repeated

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 $signs and I was not able to understand it. • Solve what? You haven't asked a question. Do you want a count of such words? Is it exactly one vowel, or possibly more? – Thomas Andrews Oct 3 '15 at 14:48 • I apologize for not being clear enough! How many 7-letter words can I form based on those criteria? And only one vowel – user276520 Oct 3 '15 at 14:53 ## 2 Answers The$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. EDIT: 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.

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