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I'm studying for my exams and stuck on this one question.

The way I'm thinking of doing this is by:

$$26^9 - \binom{26}3-\binom{26}2-\binom{26}1-\binom{26}0= 5,429,503,676,728$$

But that seems very, very wrong. Any help would be appreciated. Thanks!

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Not many in Welsh. More in English. – 1015 Apr 7 '13 at 4:15
Is 'Y' a vowel? – Marc van Leeuwen Apr 7 '13 at 4:48
up vote 5 down vote accepted

One way to see that it’s almost certainly wrong is to notice that it doesn’t in any way take into account the fact that there are $5$ vowels. On the other hand, the idea of starting with all $26^9$ strings and throwing out the ones with fewer than $3$ vowels is good. Specifically, you want to throw out the ones that have $0,1$, or $2$ vowels. (You went a step too far.)

There are $5$ vowels and $21$ consonants, so there are $21^9$ strings composed entirely of consonents $-$ i.e., with $0$ vowels.

To make a string with exactly $2$ vowels, you must choose which $2$ of the $9$ positions are to be filled with vowels, then choose vowels for those $2$ positions, and finally choose consonants for the other $7$ positions. You can do that in $\binom92\cdot5^2\cdot21^7$ ways.

How many ways are there to make a string with exactly one vowel?

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Will it be (c(9,1))(5)(21^8)? – StackPWRequirmentsAreCrazy Apr 7 '13 at 4:26
@StackPWRequirmentsAreCrazy: Looks good. – Brian M. Scott Apr 7 '13 at 4:27
Thanks so much :). Would my last step be to add the 0 vowels result, the 1 vowel result and the 2 vowels results and subtract the sum from 26^9? – StackPWRequirmentsAreCrazy Apr 7 '13 at 4:29
@StackPWRequirmentsAreCrazy: That’s exactly right. You’re welcome! – Brian M. Scott Apr 7 '13 at 4:30
Thanks once again. People like you make this website great. – StackPWRequirmentsAreCrazy Apr 7 '13 at 4:32

The strategy you used is basically right. A number of the details are not. There are $26^9$ strings. We subtract the number of "bad" ones.

There are $21^9$ all consonant ($0$ vowel) strings.

Now count the $1$ vowel strings. The location of the vowel can be picked in $\binom{9}{1}$ ways. That location can be filled with a vowel in $5^1$ ways, And the remaining $8$ locations can be filled with consonants in $21^8$ ways, for a total of $\binom{9}{1}5^121^8$.

Now we count the strings with exactly $2$ vowels. Where the vowels will go can be chosen in $\binom{9}{2}$ ways. These locations can be filled with vowels in $5^2$ ways. And for each of thsse ways, the remaining $7$ spots can be filled with consonants in $21^7$ ways.

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@StackPWRequirmentsAreCrazy: Yes, exactly. I figured it was not necessary to say that since one could see that you had that strategy in mind, except that your count of the "bads" was not right. I don;t know whether in your course they want "numbers" or expressions. To me, expressions are more informative. – André Nicolas Apr 7 '13 at 4:36
Thank you very much for your reply :). It was a huge number but I think it's correct. – StackPWRequirmentsAreCrazy Apr 7 '13 at 4:44
You are welcome. For me, a big number written out is a meaningless jumble of digits. I like structure. – André Nicolas Apr 7 '13 at 4:53

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