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

• jaus tail
In summary, the homework statement is that there are permutations with letters that are nPr where n is the total number of things, r is the number of things taken at a time, and P1 and P2 are the number of things that are identical. If there are P1 objects and P2 objects that are identical, then the permutation is n!/(P1!P2 !).
jaus tail

## Homework Statement

All permutations made with letters (a, a, b, c) taken two at a time are:

## Homework 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 !)

## 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:
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.

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

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...

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.

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.

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.

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.

Well i learned 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.

## 1. What is the total number of possible permutations when taking 2 letters from (a, a, b, c)?

The total number of possible permutations when taking 2 letters from (a, a, b, c) is 6.

## 2. How do you calculate the number of permutations in this scenario?

The number of permutations can be calculated using the formula nPr = n! / (n-r)!, where n is the total number of elements and r is the number of elements taken at a time.

## 3. Is order important in these permutations?

Yes, order is important in these permutations. For example, taking "a" first and then "b" is different from taking "b" first and then "a".

## 4. Can there be repetitions in these permutations?

Yes, there can be repetitions in these permutations. In this scenario, there are two "a" letters, so it is possible to have combinations like "aa" or "ab".

## 5. How do you account for the duplicate "a" letters in these permutations?

The duplicate "a" letters can be accounted for by dividing the total number of permutations by the factorial of the number of duplicate elements. In this case, we divide by 2! since there are two "a" letters. This gives us a final answer of 6 / 2! = 3 permutations.

• Calculus and Beyond Homework Help
Replies
6
Views
881
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Programming and Computer Science
Replies
4
Views
1K
• Precalculus Mathematics Homework Help
Replies
8
Views
799
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
684
• Calculus and Beyond Homework Help
Replies
2
Views
504
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K