How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

And by 'words' I mean just a string of alphabet - does not have to be a real world in dictionary. I do know that you need to use Inclusion/Exclusion principle which says:

$|A \cup B| = |A| + |B| - |A\cap B|$

Let U be the set of all words containing 7 letters (26 letters to the power 7) The only thing I'm not sure is what would be the sets A and B? Or do I need a third term C?

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It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).

I assume that by "$3$" or more vowels" you mean $3$ or more occurrences of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.

How many words are there with no vowels? Clearly $$21^7$$ if, as per usual convention, we agree that there are $5$ vowels.

How many words with $1$ vowel? Where the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways. And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of $$\binom{7}{1}(5)(21^6)$$

Finally, how many with $2$ vowels? The location of the vowels can be chosen in $\binom{7}{2}$ ways. Once this has been done, the actual vowels can be put into these places in $5^2$ ways. And then you can fill in the consonants in $21^5$ ways, for a total of $$\binom{7}{2}(5^2)(21^5)$$

Add up the $3$ numbers we have obtained, subtract from $26^7$.

Our argument was a little indirect. We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up. This is only a little more work than the indirect approach. But any saving of work is helpful! Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.

Remark: The calculation we have done is closely connected to the Binomial Distribution, and if you have already covered this, it may be the point of the exercise. So if you know about the Binomial Distribution, imagine the letters to be chosen at random. Then the number of patterns with $3$ or more vowels is the probability of $3$ or more vowels, multiplied by $26^7$.

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Thanks heaps! I get it now. Actually it wasn't about Binomial Distribution, but good to know! –  Arvin May 17 '11 at 13:28

I'm going to assume you are talking about the english language with letters, $L = \{ a, b, c, ... x, y, z \}$ ($|L| = 26$) and vowels, $V = \{a, e, i, o, u \}$ ($|V| = 5$). Let $C = L \setminus V$ be the set of consonants ($|C| = 21$).

Let's say you had no restrictions. You have $n$ positions ($n \ge 1$), each position has an element selected from $L$. Thus you have $|L|^n$ possible arrangements.

If you require at least $m$ vowels in your string ($m \le n$), then you will have to consider $n - m + 1$ cases.

Consider $m = n$. There is one case to consider, all of the characters are vowels and thus the total number of arrangements is $|V|^n$.

Consider $m = n - 1$. There are two cases that you will consider. The previous case and the one where all but one position is a vowel. There are $C(n,1)$ ways to position the consonant in the string. So you have $|V|^{n-1} |C| C(n,1)$ arrangements.

Consider $m = n - 2$. There are three cases. The previous two cases and the one where all but two positions are a vowel. So you have $|V|^{n-2} |C|^2 C(n, 2)^2$.

Thus, consider $m = n - k$. There are $k + 1$ cases. The previous $k$ cases and the one where all but $k$ positions are a vowel. You have $|V|^{n-k} |C|^k C(n, k)^k$.

To account for everything, you have $\sum_{i = m}^{n}{ |V|^{i} |C|^{n-i} C(n, n-i)^{n-i} }$.

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Hmm, I think I see what you're doing with this method, but it looks very complicated and I don't really understand it fully. I think the above answer is much simpler, but thanks! –  Arvin May 17 '11 at 13:31

Hint: I wouldn't use inclusion/exclusion at all. First calculate how many words of precisely 4 consonants and 3 vowels you can make. Which does "y" count as? You need to select 4 consonants, then select 3 vowels, then select an ordering of consonants/vowels. How many ways can you do each of these? Then repeat for 4,5,6, and 7 vowels.

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Yes, it turns out we weren't supposed to use Incl/Excl. I was shown another "At least" type question and we had to use Incl/Excl for that so I assumed we needed it for all "At least" type questions. Thanks! –  Arvin May 17 '11 at 13:30