I believe this is a Combinations problem

[eric12872]

If a "word" is interpreted to be a sequence of letters, how many 8 letter words with no repeated letters contain all five vowels? (Note: that y is not considered a vowel.).

Can someone teach me how to do this one?

 

[HallsofIvy]

`The vowels, since we are not including y, are "a, e, i, o, u". There are 5 vowels and 21 consonants. You want 8 letters containing exactly 5 vowels and 3 consonants. First, if we write "c" for any consonant and "v" for any vowel, how many different orders are there for "vvvvvccc"? Now there are 5 choices for the first vowel, 4 choices for the second vowel, then three choices for the third, etc. There are 5*4*3*2*1= 5! ways to choose the vowels. There are 21 choices for the first consonant, 20 choices for the second consonant and 19 choices for the third so 21*20*19= 21!/18! ways to choose the consonants.

 

[eric12872]

`The vowels, since we are not including y, are "a, e, i, o, u". There are 5 vowels and 21 consonants. You want 8 letters containing exactly 5 vowels and 3 consonants. First, if we write "c" for any consonant and "v" for any vowel, how many different orders are there for "vvvvvccc"? Now there are 5 choices for the first vowel, 4 choices for the second vowel, then three choices for the third, etc. There are 5*4*3*2*1= 5! ways to choose the vowels. There are 21 choices for the first consonant, 20 choices for the second consonant and 19 choices for the third so 21*20*19= 21!/18! ways to choose the consonants.

Ok thank you. So now for vowels, I have 120 ways to choose the vowels and 7980 ways to choose the consonants. Would I then divide the consonants by the vowels: 7980/120 = 66.5. Am I allowed to get a fraction. Did Ii go wrong somewhere?

 

[HallsofIvy]

What reasoning do you have for "dividing the consonants by the vowels"? Have you yet calculated the number of ways you can order "vvvccccc"?

(Edited)

 

[eric12872]

What reasoning do you have for "dividing the consonants by the vowels"? Have you yet calculated the number of ways you can order "vvvcccccccccc"?

Ok why did you write it like "vvvcccccccccc"? It is vvvvvccc -- I get really confused really fast with math so I need to make every bit make sense before I can move on. So how would I calculate VVVVVCCC? I don't want to be difficult, this stuff just comes a lot harder for me than probably the average poster on this forum.

 

[HallsofIvy]

You labeled this a "combinations" problem so I assumed you were taking a course involving "permutations and combinations". The number of ways you can write three "v"s and 5 "c"s is a permutation: \(\displaystyle \begin{pmatrix}8 \\ 5\end{pmatrix}= \begin{pmatrix}8 \\ 3\end{pmatrix}= \frac{8!}{5! 3!}\).


More generally, the number of ways you can permute n objects, i of which are identical and j of which are also identical, i+ j= n, is \(\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}= \frac{n!}{i! j!}\)

 

[soroban]

Hello, eric12872!

If a 'word' is interpreted to be a sequence of letters,
how many 8-letter 'words' with no repeated letters contain all five vowels?
(Note: that y is not considered a vowel.)

There are 5 vowels and 21 consonants.

We will take all 5 vowels and 3 of the 21 consonants:

. . \(\displaystyle {21\choose3} \,=\,1330\) choices.

Then the 8 letters can be arranged in \(\displaystyle 8! \,=\,40,\!320\) ways.

Therefore, there are: \(\displaystyle \:1,\!330 \times 40,\!320 \;=\;53,\!625,\!600\) eight-letter 'words'.

 

[RhondaSommer]

I have a question?

If a "word" is interpreted to be a sequence of letters, how many five-letter words with no repeated letters contain two vowels? (Note that y is not considered a vowel.)

I know that the vowels, since we are not including y, are "a, e, i, o, u". There are 5 vowels and 21 consonants. You I want 5 letters containing exactly 2 vowels and 3 consonants. If I write "c" for any consonant and "v" for any vowel, how am I supposed to know how many different orders are there for "vvcccc"?

I've read throughout the entire thread but I'm still not sure what to answer...

 

[stapel]

If a "word" is interpreted to be a sequence of letters, how many five-letter words with no repeated letters contain two vowels? (Note that y is not considered a vowel.)

I know that the vowels, since we are not including y, are "a, e, i, o, u". There are 5 vowels and 21 consonants. You I want 5 letters containing exactly 2 vowels and 3 consonants. If I write "c" for any consonant and "v" for any vowel, how am I supposed to know how many different orders are there for "vvcccc"?

I've read throughout the entire thread but I'm still not sure what to answer...

What question do you have regarding the posted solution? Please be specific. Thank you!