Question

English has $6$ letters that can be vowels. This includes Y, which can be either a consonant or a vowel; for this problem, we'll consider Y a vowel.
The other $20$ English letters are always consonants.

How many two-letter "words" can we make from these letters if we are required to use at least one vowel? (We aren't limited to words that have an actual meaning in English. Thus, for this problem, we'll include nonsense "words" like AA, QO, XY, and UZ.)

Answer 1

Answer:

276

Step-by-step explanation:

There are three types of words we're allowed to form: vowel-consonant, vowel-vowel, and consonant-vowel. We treat these as separate cases.

The first type of word (vowel-consonant) can be formed in $6\cdot 20=120$ ways, since we have $6$ choices for the first letter and $20$ choices for the second letter.

The second type of word (vowel-vowel) can be formed in $6\cdot 6 = 36$ ways.

The third type of word (consonant-vowel) can be formed in $20\cdot 6 = 120$ ways.

Adding up all these possibilities, we have $120+36+120 = \boxed{276}$ possible two-letter words with at least one vowel.

Solution #2

If we ignore the requirement to have at least one vowel, then we have 26 choices for each letter, making $26\cdot 26 = 676$ two-letter "words". Now we eliminate the $20\cdot 20 = 400$ words whose letters are both consonants. This leaves $676-400=\boxed{276}$ words with at least one vowel.

Answer 2

If the first letter is a vowel, there are 6*26 = 156 possibilities.
If the second letter is a vowel, there are 26*6 = 156 possibilities.

The sum of these numbers is 2*156 = 312, but that counts double-vowel words twice. There are 36 of those.

The number of possible unique 2-letter pairs is 312 - 36 = 276.