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I considered two cases.

Case 1 (2 vowels): Pick 2 vowels $\binom{5 + 2 -1}{2}$, then pick 4 consonants $\binom{21 + 4 -1}{4}$, then order them $6!$.

Case 2 (4 vowels): Pick 4 vowels $\binom{5 + 4 -1}{4}$, then pick 2 consonants $\binom{21 + 2 -1}{2}$, then order them $6!$.

Total: $\binom{5 + 2 -1}{2} \binom{21 + 4 -1}{4} 6! +\binom{5 + 4 -1}{4} \binom{21 + 2 -1}{2} 6!$

But then I realized I couldn't do the factorial step because I could have the same vowels/consonants appearing more than once so I would be over counting their order. How can I clear this up?

Is there an alternative way I can go about this?

Thanks!

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Please don't add "NOT HOMEWORK" to your title. We have sort of an honor system here: if you don't tag a question as homework, we generally assume it is not homework. There are of course exceptions (i.e. questions which just state an exercise and show no work), but in those cases saying a question isn't homework doesn't convince anyone. –  Alex Becker Dec 16 '13 at 3:33
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+1 for showing your work. Even if it were homework, you would get a good reception here because of this. –  Ross Millikan Dec 16 '13 at 3:35
    
I know, but it seems like every time I don't mention it, someone says, "Were not here to do your homework for you! well still give you help but make sure to add the "Homework" tag" This isn't homework I'm studying for a final tomorrow night. –  mharris7190 Dec 16 '13 at 3:35
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2 Answers 2

up vote 5 down vote accepted

We can get around it as follows for the two vowel case:

  1. Represent the word as six blank spots: _ _ _ _ _ _.

  2. Choose two spots for the vowels: ${6 \choose 2}$.

  3. Since order matters, we can fill those two spots in $5^2$ ways.

  4. Since order matters, we can fill the remaining consonants in $21^4$ ways.

For a total of ${6 \choose 2} 5^2 21^4=72930375$ ways.

A similar argument applies to the four vowel case. I get a final total of $77064750$ different words with either exactly two or exactly four vowels.

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Take it from the slot's perspective. It must choose one of the $5$ vowels. Since there are $2$ slots, there are $2$ factors of $5$, or $5^2$. The difference from ball-in-box problem is that multiple balls can go into each box, but slots cannot have multiple vowels. We can reduce it to a ball-in-box problem by saying the slots are the balls and the vowels are the boxes. Each slot picks a vowel to "go into". Notice that each vowel can "have" more than one slot, since there can be repeats--just like the ball-in-box problem! But now since the roles are reversed, the answer is $5^2$, not $2^5$. –  Eric Thoma Dec 16 '13 at 4:18
    
@mharris7190 If your follow-up is not already answered, I have answered above. For others, the follow-up was about viewing the problem from the perspective of putting distinguishable balls into distinguishable boxes without exclusion--a very good question. –  Eric Thoma Dec 16 '13 at 4:19
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You have made good progress. I will focus on case 1, which shows the issues. When you pick two vowels, you might pick two the same. You are correct that there are $15$ combinations of two vowels where they might be the same. ${5 \choose 2} = 10$ of those have two distinct vowels and ${5 \choose 1}=5$ of those have two vowels the same. If you want to compute the number of words with a specific set of four consonants, two of which are the same, you could do ${6 \choose 2}$ ways to pick the positions of the vowels, $\frac {4!}{2!}$ ways to order the consonants and $2$ or $1$ ways to order the vowels depending on whether they were the same. The total is $12 (10 \cdot 2 + 5 \cdot 1)=300$

I will leave this, as I think it shows a way to think about other problems. But Eric Thoma has found a very simple solution to this one.

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