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We would like to construct $3$ letter strings from the english alphabet ($26$ letters: $21$ consonants and $5$ vowels).

a. How many ways can you construct strings with no vowel?

b. How many ways can you construct strings with at least one vowel?

This is what I did for both of them... however I think at least one of them is incorrect.

a. Ways with no vowel = $21^3$ since for each letter, we have $21$ options.

b. Ways with at least one vowel = total ways with no restrictions - ways with no vowel
= $26^3 - 21^3$.

However, manually computing it:
Ways with at least one vowel = ways with one vowel + ways with two vowels + ways with three vowels = $\binom{5}{1} \times \binom{21}{2} \times 3! + \binom{5}{2} \times \binom{21}{1} \times 3! + \binom{5}{3} \times \binom{21}{0} \times 3!$

which doesn't equal to the other way.

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1 Answer 1

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In the second case, you are not taking into account repeats. For example your way of calculating the options for making a word with 2 consonants only allows 2 different consonants, not repeats as your definition of a string would indicate. Calculating the number of cases with one vowel should be $5*21^2*3$

So overall, it should be $5*21^2*3+5^2*21*3+5^3=8315$

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  • $\begingroup$ I thought that if we had $21\times 21$, we have already "arranged" the consonants, so there is no need for a $3!$ there is there? $\endgroup$
    – Natash1
    Sep 27, 2017 at 13:29
  • $\begingroup$ Just as in part a, I didn't times it by $3!$ because I think $21\times 21\times 21$ accounts for the arrangements? $\endgroup$
    – Natash1
    Sep 27, 2017 at 13:30
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    $\begingroup$ Apologies, but you still need to multiply it by 3, because your vowel position could be in the middle of the consonants, on the left or right of two consonants. I have made the necessary edit $\endgroup$
    – Harry Alli
    Sep 27, 2017 at 13:32
  • $\begingroup$ And in the second case, we look at it from the consonant perspective: the consonant can be in the middle, or the left and right of vowels? $\endgroup$
    – Natash1
    Sep 27, 2017 at 13:40
  • $\begingroup$ Yeah, that is the case $\endgroup$
    – Harry Alli
    Sep 27, 2017 at 13:42

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