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?


  • 1
    $\begingroup$ +1 for showing your work. Even if it were homework, you would get a good reception here because of this. $\endgroup$ Dec 16, 2013 at 3:35

2 Answers 2


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.

  • $\begingroup$ 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 $2^5$. $\endgroup$
    – Eric Thoma
    Dec 16, 2013 at 4:18
  • $\begingroup$ @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. $\endgroup$
    – Eric Thoma
    Dec 16, 2013 at 4:19

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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