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

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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: $5^2 \cdot 21^6$.

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  • $\begingroup$ The Separation Technique involves arranging objects such that there is a space between each one, in this case (c is consonant): _c_c_c_c_c_c_ We then pick the necessary number of spots for the non-adjacent objects, in this case: C(7,2) and fill those spots, in this case: 5^2 $\endgroup$ – B.K. Apr 17 '16 at 18:39
  • $\begingroup$ I actually made a mistake in my solution... just edited it. $\endgroup$ – B.K. Apr 17 '16 at 18:44
  • $\begingroup$ You indeed get the same answer, I see. $\endgroup$ – Henno Brandsma Apr 17 '16 at 18:50
  • $\begingroup$ I am glad, though, that you showed me your way, as it made things a bit more clear. It's the same as the first part, except now we just take out one of the possibilities where they're next to each other. So, instead C(8,2), it's C(7,2), which is the same as C(8,2) - 7. Thank you for your answer. $\endgroup$ – B.K. Apr 17 '16 at 18:56
  • $\begingroup$ Glad to be able to help! $\endgroup$ – Henno Brandsma Apr 17 '16 at 18:56

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