# Homework Help: 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

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.