Solving using permutation and combination

1. Jul 13, 2015

rajeshmarndi

1. The problem statement, all variables and given/known data
From 5 consonants and 4 vowels, how many words can be formed consisting of 3 consonants and 2 vowels

The book solved it using combination
C(5,3) * C(4,2) * 5! = 7200

i.e I understand the 1st two term give the unique 5 words that can be formed with 5 consonants and 4 vowels and multiplying with 5! gives the permutation of all the 5 letters words
2. Relevant equations

3. The attempt at a solution
I tried on my own and used permutation
P(5,3) * P(4,2) = 720

I understand the book solution but then where am I wrong here?

2. Jul 13, 2015

PeroK

Your method counts the vowels and consonants separately and doesn't take into account that they can then be mixed up. E.g. you could have:

$C_1C_5C_3$ $V_4V_2$

And there are then 10 ways to put these together into words with the vowels and consonants in that order.

3. Jul 13, 2015

rajeshmarndi

Yes thanks, i.e C1C2C3 can be as such C1C2C3_ _ , C1 _ C2C3 _ , C1 _ _ C2C3 and so on.

4. Jul 13, 2015

rajeshmarndi

Just to add more, P(5,3) give C5C3C1 is counted once but it can be placed more than once in the 5 lettered words which I didn't counted as, C5_C3C1_ , C5_ _C3C1 and others similarly.

5. Jul 14, 2015

HallsofIvy

Staff Emeritus
Are these 5 distinct consonants and 4 distinct vowels? That's an important piece of information.

6. Jul 14, 2015

rajeshmarndi

Yes I have been messing around, I had realized it lately.

The permutation of P(5,3) = 60 is at , say, 1st three position and along with it permutation of P(4,2) = 12 . This doesn't change location, say, of P(5,3) which can be placed in C(5,3) = 10 ways in the 5 lettered words, along with the permutation of P(4,2).

It then becomes P(5,3) * P(4,2) * C(5,3) = 60 * 12 * 10 = 7200.

Sometimes it becomes so hard to see how permutation/combination actually give the answer.