Combinatorics

[Roger.Robert]

Consider all strings of length l ∈ {1, 2, 3} consisting of letters from the set {a, b, c, d, e}.
How many of these strings have an even amount of vowels? (a and e are vowels).

Here's how i tried to solve it :
5 strings of length 1 . - 3 strings with an even amount of vowels
5!/(5-2)!=20 strings of length 2 - 8 strings with an even amount of vowels.
5!/(5-3)!=60 strings of length 3 -24 strings with an even amount of vowels.
In total 85 strings. And 35 strings with an even amount of vowels.But it's wrong, and idk why.(79 is the answer in the book)

 

[Dr.Peterson]

Consider all strings of length l ∈ {1, 2, 3} consisting of letters from the set {a, b, c, d, e}.
How many of these strings have an even amount of vowels? (a and e are vowels).

Since you didn't show any work or tell us where you are struggling, I have to assume it is at the start.

I would start by interpreting the problem. First, we have to decide whether repetitions are allowed; I will assume they are (as that is what "strings" typically means. The length of the string can be either 1, 2, or 3. So we have three questions, and we'll add up the answers:

• How many "strings" of length 1 are there, with NO vowels (since the number of vowels can be either 0 or 1, and 0 is even)? That is, how many single letters that are not vowels do we have to choose from?
• How many strings of length 2 are there consisting either of no vowels or 2 vowels?
• How many strings of length 3 are there, containing either no vowels or 2 vowels?

Give it a try, so we can see how close you are to an answer, and can know where you need help.

 

[Roger.Robert]

Since you didn't show any work or tell us where you are struggling, I have to assume it is at the start.

I would start by interpreting the problem. First, we have to decide whether repetitions are allowed; I will assume they are (as that is what "strings" typically means. The length of the string can be either 1, 2, or 3. So we have three questions, and we'll add up the answers:

• How many "strings" of length 1 are there, with NO vowels (since the number of vowels can be either 0 or 1, and 0 is even)? That is, how many single letters that are not vowels do we have to choose from?
• How many strings of length 2 are there consisting either of no vowels or 2 vowels?
• How many strings of length 3 are there, containing either no vowels or 2 vowels?

Give it a try, so we can see how close you are to an answer, and can know where you need help.

Here's how i tried to solve it :
5 strings of length 1 . - 3 strings with an even amount of vowels
5!/(5-2)!=20 strings of length 2 - 8 strings with an even amount of vowels.
5!/(5-3)!=60 strings of length 3 -24 strings with an even amount of vowels.
In total 85 strings. And 35 strings with an even amount of vowels.But it's wrong, and idk why.(79 is the answer in the book)

 

[Dr.Peterson]

Here's how i tried to solve it :
5 strings of length 1 . - 3 strings with an even amount of vowels
5!/(5-2)!=20 strings of length 2 - 8 strings with an even amount of vowels.
5!/(5-3)!=60 strings of length 3 -24 strings with an even amount of vowels.
In total 85 strings. And 35 strings with an even amount of vowels.But it's wrong, and idk why.(79 is the answer in the book)

You're using permutations. Your answers are correct (though you didn't show how you got them) if permutations are what they want -- good work!

But recall that my first observation was to ask whether repetitions are allowed, and conclude that most likely the answer is yes, from my experience with the word "string". That means, for example, that we would include the string bbb, which is not a permutation.

You've made the opposite assumption, did correct work, and found the "wrong" answer, so that must not be their intention. (In a classroom setting, I would ask the teacher to confirm my interpretation or correct it.)

Try again, interpreting it my way, and see if you get 79.