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The english alphabet contains $21$ consonants and $5$ vowels. How many strings of $6$ lowercase letters of the English alphabet contain

  • a) exactly 1 vowel
  • b) exactly 2 vowels
  • c) at least one vowel
  • d) at least two vowels
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Exactly $1$ vowel: The location of the vowel can be chosen in $\binom{6}{1}$ ways. (Of course this is $6$, but we are trying to use a technique that works more generally.)

For each choice of location, the location can be filled with a vowel in $5$ ways.

That leaves $5$ empty locations, which can be filled with consonants in $21^5$ ways. That gives a total of $$\binom{6}{1}(5)(21^5).$$

Exactly $2$ vowels: The location of the vowels can be chosen in $\binom{6}{2}$ ways.

For each choice of locations, the locations can be filled with vowels in $5^2$ ways.

That leaves $4$ empty locations, which can be filled with consonants in $21^4$ ways. That gives a total of $$\binom{6}{2}(5^2)(21^4).$$

At least $1$ vowel: There are $26^6$ $6$-letter "words." And there are $21^6$ all consonant words. So there are $26^6-21^6$ words that have at least one vowel.

There are other ways of counting this, but they are less efficient.

At least $2$ vowels: Again, there are various ways of counting. An efficient way is to count the words that have $0$ vowels or $1$ vowels, and subtract from the total number of words. This approach lets us recycle previous results, and recycling is a virtue.

So take the total number $26^6$ of words, and subtract the $21^6$ all consonant words, and the number of $1$-vowel words we already calculated.

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    $\begingroup$ I realize this answer is over two years old, but I feel the need to clarify, since I was initially confused: when @André Nicolas says "the number of 1-vowel words we already calculated" in the last sentence, he's referring to the number of "exactly 1 vowel" strings, not "at least 1 vowel" strings. $\endgroup$ – Justin Lardinois Jan 27 '15 at 7:42
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    $\begingroup$ @JustinLardinois: Yes, the $1$-vowel words mentioned in the final sentence refers to the exactly $1$-vowel words. I would insert the word exactly, except that editing bumps an old question to the front page, OK for a major change but somewhat frowned on otherwise. $\endgroup$ – André Nicolas Jan 27 '15 at 9:46

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