# Permutations of taking 2 letters from (a, a, b, c)

1. Feb 6, 2017

### jaus tail

1. The problem statement, all variables and given/known data
All permutations made with letters (a, a, b, c) taken two at a time are:
2. Relevant equations
Permutation is nPr where n is total number of things, r is number of things taken at a time.
If P1 objects are identical, P2 objects are identical such that P1 + P2 = n
then permutation is
n!/(P1!P2 !)

3. The attempt at a solution
Here letters are (a, a, b, c)
Four letters. n = 4.
Taking 2 at a time. means r = 2

But two letters repeat so P1 = 2.
letter b comes once. so P2 = 1
Letter c comes once so P3 = 1

Permutation is: 4!/(2!)(1!)(1!) = 4 * 3*2/(2) = 12

But if i arrange the letters I get:
(a, a, b, c) taking 2 letters at a time:

1. aa
2. ab
3. ba
4. ac
5. ca
6. bc
7. cb
I get 7 combinations. Where am I wrong?
Whereas by formula I get answer as 12. (above in italics)

PS ( I know I'm posting a lot. But i have a big exam of math and electronics on 11th feb and thus am revising all topics together. )

Last edited: Feb 6, 2017
2. Feb 6, 2017

### Orodruin

Staff Emeritus
You are applying an argumentation that does not apply. $n!/(p_1!\ldots p_k!)$ has no direct relation to what you want to compute, it is the number of different unordered ways of placing $n$ objects into $k$ bins such that bin $k$ contains $p_k$ objects.

3. Feb 6, 2017

### Orodruin

Staff Emeritus
Also, 4!/(2!) = 4*3*2/2 = 12.

4. Feb 6, 2017

### jaus tail

Hmm... thanks for pointing out my mistake. I've rectified it now.
Though honestly i'm still not clear when to use the formula and when to not.
Will read more examples from book and hope that helps..

Thanks...

5. Feb 6, 2017

### Orodruin

Staff Emeritus
12 would be the number of possibilities if your letters were distict. The boxes would be 2 boxes with a single letter (first and second letter, respectively) and one box with two letters (the unused letters). Now, your letters are not distinct so you will need to account fore double counting duplicates.

6. Feb 6, 2017

### jaus tail

Can I put the first 2 letters in separate boxes?
Question says permutations with 2 letters taken at a time from (a, a, b, c).
So i got 2 boxes and i need to put letters from here.
Case can be:
aa, ab, ba, ac, ca, bc, cb

Is there some formula for this?

I tried another case:
letters (a, a, b, c, d)
Permutations with 2 letters is:
aa
ab ba ac ca ad da bc cb bd db cd dc
Total = 13.
But i couldn't come up with any formula.

7. Feb 6, 2017

### Orodruin

Staff Emeritus
I think you should stop focusing on having a formula and instead think about how to reason in order to arrive at the correct result. It will help you in the long run.

8. Feb 6, 2017

### jaus tail

Yeah true. I made a big table with 4 and 5 letters and different repeating values. Wasn't able to derive a formula though.
Thanks.

9. Feb 6, 2017

### jaus tail

Well i learnt one thing that the multi set equation (of dividing with P ! when I is number of identical objects) can be used only when all objects are taken.
So some good thing came out of this exercise.