1. ## Combinatory

If a word has 5 letters picked out of the whole alphabet (26 letters), what percent does not contain any vowels repeated?
I am having trouble with this problem. I would like to know what is needed to be done in order to solve it.

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3. ## Re: Combinatory

Originally Posted by Yeison
If a word has 5 letters picked out of the whole alphabet (26 letters), what percent does not contain any vowels repeated? I would like to know what is needed to be done in order to solve it.
Your word string can contain anywhere from zero to five vowels no one of which is repeated,
Combinations are $\displaystyle\binom{N}{k}=\frac{N!}{k!(N-k)!}$. Permutations are $\displaystyle \mathcal{P}^N _k=\frac{N!}{(N-k)!}$

The answer is: $\displaystyle\sum\limits_{k = 0}^5 {\binom{5}{k} \cdot \mathcal{P}^5_k \cdot {{\left( {21} \right)}^{5 - k}}}$

The explanation is: there are k=0 to 5 vowels, there 5 places we place the vowels, then we permute the vowels, the 5-k remaining places can be filled with any of 21 consonants.

4. ## Re: Combinatory

Hello, Yeison!

Would you clarify the wording of this problem?

If a word has 5 letters picked out of the whole alphabet (26 letters),
what percent does not contain any vowels repeated?

You used the term "word".
Does this mean that the order of the letters in considered?

If the five letters are chosen without replacement,
. . no letter is repeated.
So, the choices are made with replacement?

5. ## Re: Combinatory

I think it should be pretty clear that a word does consider the order of the letters.

He states that only vowels aren't to be repeated. It should be pretty clear that there is no such restriction on consonants and thus sampling of consonants is done with replacement.

6. ## Re: Combinatory

I agree . . . all of that is "pretty clear".

But from past experiences, I just wanted to make sure.