[Math] Exactly two vowels in an eight-letter string.

combinatorics

Could someone confirm my combinatorics solutions for this question?

Part 1

How many eight letter strings of letters contain exactly two vowels?

Solution:

  1. Choose two spots out of eight possible for the two vowels, order does not matter — $C(8,2)$.
  2. Pick a vowel for each spot. There are two spots, five vowels in the alphabet and "no repeats" condition was not specified, so there are $5^2$ choices.
  3. Pick the remaining six consonants, which is $21^6$, since there are $21$ consonants and six spots left.

Answer: $C(8,2) \cdot 5^2 \cdot 21^6$

Part 2

How many eight letter strings of letters contain exactly two vowels if the two vowels cannot be adjacent?

Solution:

  1. Using the Separation Technique, space out and place the six possible consonants, creating seven possible positions for the two vowels — $21^6$.
  2. Out of the seven spacer spots, pick two to be used for the two vowels — $C(7,2)$.
  3. There are five choices per spot and "no repeats" restriction was not specified — $5^2$.

Answer: $21^6 \cdot C(7,2) \cdot 5^2$

Best Answer

I agree with the first one completely.

I'm not sure what the Separation Technique is.

I would count the positions for the vowels that are non-adjacent. We have 7 pairs of places that are adjacent: 12, 23,..., 78. So we have $\binom{8}{2} - 7$ many position pairs for the vowels.

The rest of the counting is the same: $5^2 \cdot 21^6$.