1. ## Permutations/Combinations help

Hey all Im having trouble with this 1 problem, hope you guys can help me through it and understand it.

The English alphabet contains 21 consonants and five vowels. How many strings of six lowercase letters of the English alphabet contain:

a)exactly 1 vowel, the answer is 122523030, but I don't know how they got the answer. I did 21*21*21*21*21*5, but didn't get the same answer.

b)exactly two vowels, I did 21*21*21*21*5*5 but is wrong the answer is 72930375

c)at least one vowel: I don't quite understand the concept of calculating problems with at least or at most.

d) at most 2 vowels

2. For the first one you are close: $\displaystyle {21}^5{5 \choose 1}{6 \choose 1}$, you choose one vowel from five, but there are six places to insert it in the string.

$\displaystyle {21}^6$ is the number of ways to have no vowels so $\displaystyle {26}^6 - {21}^6$ is the number of ways to have at least one.

3. Ok, so it was 21^5 * 5 vowels * 6 places to place them
which is 21^5*30

but for the next problem for exactly 2 vowels in it.
the answer is 72930375 but I didn't get it.

isn't it 21^4 * 5 vowels * 6 places * 5 vowels * 5 places?
in other words isn't it 21^4*750?

4. Originally Posted by ff4930
isn't it 21^4 * 5 vowels * 6 places * 5 vowels * 5 places? in other words isn't it 21^4*750?
Not quite.
Try $\displaystyle {6 \choose 2}{5}^2{21}^4$
Choose two places to put the vowels.

The last one is:
$\displaystyle \sum\limits_{k = 0}^2 {6 \choose k} \left( 5 \right)^k \left( {21} \right)^{6 - k}$

5. ahh, I got it. Its combination of 2 out of 6 * 5^2 then * 21^4.

Sorry to be a bother but, problems like at least one vowel or at most two vowels. How would I approach this? I don't seem to get it, if it is at least one vowel then if I find exactly one vowel wouldn't that fulfill the requirement of least one vowel?

6. I answered both of those above:

AT LEAST ONE
$\displaystyle {21}^6$ is the number of ways to have no vowels so $\displaystyle {26}^6 - {21}^6$ is the number of ways to have at least one.

AT MOST TWO
$\displaystyle \sum\limits_{k = 0}^2 {6 \choose k}\left( 5 \right)^k \left( {21} \right)^{6 - k}$