# Permutations with possibly repeating letters

• poonintoon
In summary, the number of permutations when there are repeating letters in a word can be calculated by dividing the total number of possible arrangements by the factorial of the number of times each letter is repeated. The order of the repeating letters does not affect the number of permutations. This concept is useful in genetics, cryptography, and computer science. The difference between permutations and combinations with repeating letters is the consideration of order. The number of permutations with repeating letters will always be equal to or less than the number of permutations without repeating letters.
poonintoon

## Homework Statement

Find the number of distinct three letter words obtainable from the letters A,B,C,D,E in which E may occur 0, 1 or 2 times but the rest may occur only once.

## Homework Equations

Number of combinations when picking r objects from n possibly objects in which one object is repeated m times
N = n! / (m! (n-r)!

## The Attempt at a Solution

I have never had a good instinct for permutations, I can plug in a formula to answer basic questions but never had a good enough understanding for it to solve more advanced problems (This is why I have dug out my old maths textbook to try and correct this )

One thing that is especially confusing for me is this.
With one E there are 5P3 = 5!/2! = 60 combinations (but this already includes ABC, ABD, ... etc i.e. all of the possible combinations with zero E's ?).
If I found the number of permutations for 3 letter words with 2E's [6!/(2!3!)] shouldn't this also include ABE etc i.e. all the permutations with 1E? Yet the answer is 60 so clearly it doesn't. I think if I could resolve my misunderstanding if this part the fog would lift for me.

Please let me know if anything is unclear and I will make some edits

poonintoon said:

## Homework Statement

Find the number of distinct three letter words obtainable from the letters A,B,C,D,E in which E may occur 0, 1 or 2 times but the rest may occur only once.

## Homework Equations

Number of combinations when picking r objects from n possibly objects in which one object is repeated m times
N = n! / (m! (n-r)!

## The Attempt at a Solution

I have never had a good instinct for permutations, I can plug in a formula to answer basic questions but never had a good enough understanding for it to solve more advanced problems (This is why I have dug out my old maths textbook to try and correct this )

One thing that is especially confusing for me is this.
With one E there are 5P3 = 5!/2! = 60 combinations (but this already includes ABC, ABD, ... etc i.e. all of the possible combinations with zero E's ?).
If I found the number of permutations for 3 letter words with 2E's [6!/(2!3!)] shouldn't this also include ABE etc i.e. all the permutations with 1E? Yet the answer is 60 so clearly it doesn't. I think if I could resolve my misunderstanding if this part the fog would lift for me.

Please let me know if anything is unclear and I will make some edits

To build your understanding, I suggest you try this problem without using the combination or permutation formulas. Try to count them more directly and find your own shortcuts.

Hint: Calculate separately the cases of 0, 1 or 2 E's.

poonintoon said:
60 combinations (but this already includes ABC, ABD, ... etc i.e. all of the possible combinations with zero E's ?).
Yes. 0 or 1 Es can be handled together because that makes E behave the same as the other letters.
poonintoon said:
If I found the number of permutations for 3 letter words with 2E's [6!/(2!3!)]
Exactly 2 Es? I make it rather less.
poonintoon said:
shouldn't this also include ABE etc
I think you are asking whether the way you arrived at 6!/(2!3!) includes ABE etc. I cannot tell since you do not explain it.

haruspex said:
Yes. 0 or 1 Es can be handled together because that makes E behave the same as the other letters.

Exactly 2 Es? I make it rather less.

I think you are asking whether the way you arrived at 6!/(2!3!) includes ABE etc. I cannot tell since you do not explain it.
Thank you, yes you have understood me correctly! That is my question. I thought that 6!/(2!3!) was basically equivalent to finding distinct permutations of ABCDEE (which would include ABE etc but clearly not). What does 6!/(2!3!) tell you the number of?

poonintoon said:
Thank you, yes you have understood me correctly! That is my question. I thought that 6!/(2!3!) was basically equivalent to finding distinct permutations of ABCDEE (which would include ABE etc but clearly not). What does 6!/(2!3!) tell you the number of?
You correctly found 60 with 0 or 1 Es. 6!/3! (=120) would count words with up to 2 Es if we regarded the 2 Es as distinct. You divided by 2 because the 2 Es are interchangeable. That gives you the number with exactly 2 Es plus half the number with 0 or 1 Es (totalling 60 again). So... how many have exactly 2 Es?

## 1. How do you calculate the number of permutations when there are repeating letters in a word?

When there are repeating letters in a word, the number of permutations can be calculated by dividing the total number of possible arrangements by the factorial of the number of times each letter is repeated. For example, if a word has 4 letters with 2 repeating letters, the number of permutations would be 4! / (2!) = 12.

## 2. Can the number of permutations change if the order of the repeating letters is changed?

No, the number of permutations remains the same regardless of the order of the repeating letters. This is because the repeating letters are considered indistinguishable in the word and therefore do not affect the total number of permutations.

## 3. How is the concept of permutations with repeating letters useful in real life?

The concept of permutations with repeating letters is useful in various fields such as genetics, cryptography, and computer science. In genetics, it can be used to calculate the number of possible combinations of genes in a population. In cryptography, it can be used to create strong passwords with repeating characters. In computer science, it can be used to optimize algorithms and data structures.

## 4. What is the difference between permutations with possibly repeating letters and combinations with repeating letters?

The main difference between permutations with possibly repeating letters and combinations with repeating letters is the order of the elements. Permutations refer to the different ways in which elements can be arranged in a specific order, while combinations refer to the different ways in which elements can be grouped without considering the order.

## 5. Can the number of permutations with repeating letters be greater than the number of permutations without repeating letters?

No, the number of permutations with repeating letters will always be equal to or less than the number of permutations without repeating letters. This is because adding repeating letters will not increase the total number of arrangements, but rather decrease it due to the decrease in distinguishable elements.

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