1. permutation help

hi, im currently student in vietnam

i have 2 permutation questions

1. how many words consisitng of 3 vowel and 2 consonants can be formed from the letters of the word "columbians"

2. a presidnet secretry and treasurer are to be named from 5 men and 3 women
how many permutations are there if one man and one women must be picked

heres how much i have done

1) you have 4 vowels and 6 consonants and 10 letters total

you need a 5 letter permutation

so you have 4 x 3 x 2 x 6 x 5

but if you do this, will you always have the order in 3 vowels and 2 consonants? what do i multiply this by to mix it up?

keep in mind my total permutations without limits is 10!/5!

2) so i do the same thing as above, i know 5 men, 3 women and a total of 8 ppl

5 x 3 x 6

but wont the man always be president? again, how do i mix this up??

sorry if my english not very good

2. Originally Posted by gogeta
1. how many words consisitng of 3 vowel and 2 consonants can be formed from the letters of the word "columbians"
Since you did not say, we assume the letters cannot be repeated.
$\dbinom{4}{3}\dbinom{6}{2}(5!)$
Choose the vowels, choose the two consonants, then arrange the five.

3. Originally Posted by Plato
Since you did not say, we assume the letters cannot be repeated.
$\dbinom{4}{3}\dbinom{6}{2}(5!)$
Choose the vowels, choose the two consonants, then arrange the five.

P(10,5) which is if I picked 5 random letters from the 10 in COLUMBIANS

4. You need to learn to do calculations:
$\dbinom{4}{3}\dbinom{6}{2}(5!)=7200$

5. Isn't this a permutation question because the order that you place the letters in matters...?

6. im unclears as to why you use combinations instead of permutation

7. Originally Posted by gogeta
im unclears as to why you use combinations instead of permutation
You must first choose three vowels $\dbinom{4}{3}=4$
Then choose the two consonants $\dbinom{6}{2}=15$
Now you have five to arrange.
This is a mixed problem.

8. Originally Posted by Plato
You must first choose three vowels $\dbinom{4}{3}=4$
Then choose the two consonants $\dbinom{6}{2}=15$
Now you have five to arrange.
This is a mixed problem.
thank you i see now that it doesn't matter which vowels you pick, b/c you mix them up by doing 5!

9. how would i be able to do the 2nd question, do i multiply by 3!

but then that would be more than P(18,3)

10. Originally Posted by gogeta
how would i be able to do the 2nd question
I will give the answer and expect you to explain it.
$^N\mathcal{P}_k$ is the permutation of N taken k at a time.
ANSWER : $^8\mathcal{P}_3~-^5\mathcal{P}_3-~^3\mathcal{P}_3$.
WHY?

11. Originally Posted by Plato
I will give the answer and expect you to explain it.
$^N\mathcal{P}_k$ is the permutation of N taken k at a time.
ANSWER : $^8\mathcal{P}_3~-^5\mathcal{P}_3-~^3\mathcal{P}_3$.
WHY?
8 permutate 3 is all possible

5P3 is all men
3P3 is all women

left with all other combos since we have subtracted all the ones that are not 2men1female or 2female1man

12. Exactly.

13. Originally Posted by Plato
Exactly.
just out of curiousity is it possible to do this question the way I have originally tried to do it in my opening post?