permutation help

Jul 2010
8
0
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
 

Plato

MHF Helper
Aug 2006
22,508
8,664
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.
\(\displaystyle \dbinom{4}{3}\dbinom{6}{2}(5!)\)
Choose the vowels, choose the two consonants, then arrange the five.
 
Jul 2010
8
0
Since you did not say, we assume the letters cannot be repeated.
\(\displaystyle \dbinom{4}{3}\dbinom{6}{2}(5!)\)
Choose the vowels, choose the two consonants, then arrange the five.
yeah but your answer is 86400, which is greater than

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

Plato

MHF Helper
Aug 2006
22,508
8,664
You need to learn to do calculations:
\(\displaystyle \dbinom{4}{3}\dbinom{6}{2}(5!)=7200\)
 
Jul 2010
8
0
Isn't this a permutation question because the order that you place the letters in matters...?
 
Jul 2010
8
0
im unclears as to why you use combinations instead of permutation
 

Plato

MHF Helper
Aug 2006
22,508
8,664
im unclears as to why you use combinations instead of permutation
You must first choose three vowels \(\displaystyle \dbinom{4}{3}=4 \)
Then choose the two consonants \(\displaystyle \dbinom{6}{2}=15 \)
Now you have five to arrange.
This is a mixed problem.
 
Jul 2010
8
0
You must first choose three vowels \(\displaystyle \dbinom{4}{3}=4 \)
Then choose the two consonants \(\displaystyle \dbinom{6}{2}=15 \)
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!
 
Jul 2010
8
0
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)
 

Plato

MHF Helper
Aug 2006
22,508
8,664
how would i be able to do the 2nd question
I will give the answer and expect you to explain it.
\(\displaystyle ^N\mathcal{P}_k\) is the permutation of N taken k at a time.
ANSWER : \(\displaystyle ^8\mathcal{P}_3~-^5\mathcal{P}_3-~^3\mathcal{P}_3 \).
WHY?