# Probability that two letters are repeated

Gold Member

## Homework Statement

A word of 6 letters is formed from a set of 16 different letters of English alphabet(with replacement). Find out the probability that exactly two letters are repeated.

## The Attempt at a Solution

Total possible words = 26C16 . 16^6
Now there are 16C2 ways to choose two letters. Letter 1 can occur atleast twice and atmost 4 times. Same is the case with letter 2. So there arises 5 cases by which repetitions can occur.

Total words possible in these cases are (6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)

∴n(P) = 26C16.16C2.(6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)
P=n(S)/n(P)
But the answer which I get is incorrect.

Mentor
so you can choose 16 letters for the first letter of the word and 16 for the second...

so the total combinations are 16^6

You could look at your problem from the reverse sense of what is the probability that no letters are repeated and then subtract it from 100%. right?

Gold Member
2021 Award
You could look at your problem from the reverse sense of what is the probability that no letters are repeated and then subtract it from 100%. right?

No, this would not work. Re-read the problem statement.

Mentor
No, this would not work. Re-read the problem statement.

You're right I overlooked that part. Close but no cigar.

Homework Helper
Gold Member

## Homework Statement

A word of 6 letters is formed from a set of 16 different letters of English alphabet(with replacement). Find out the probability that exactly two letters are repeated.

## The Attempt at a Solution

Total possible words = 26C16 . 16^6
Now there are 16C2 ways to choose two letters. Letter 1 can occur atleast twice and atmost 4 times. Same is the case with letter 2. So there arises 5 cases by which repetitions can occur.

Total words possible in these cases are (6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)

∴n(P) = 26C16.16C2.(6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)
P=n(S)/n(P)
But the answer which I get is incorrect.
It is a fixed 16 letters from the alphabet. What other letters happen to exist in the alphabet are irrelevant, so there should be no reference to 26.
If some letters are repeated then there will be fewer than 6 distinct letters chosen, so references to 6C2 are also suspect.

Gold Member
It is a fixed 16 letters from the alphabet. What other letters happen to exist in the alphabet are irrelevant, so there should be no reference to 26.
If some letters are repeated then there will be fewer than 6 distinct letters chosen, so references to 6C2 are also suspect.

By removing the 26C16 and 6C2 part I am very close to the answer. The numerator of my answer is 18090 whereas that of the answer given in my book is 18080. My denominator is correct. So I think there is a possibility that the answer in my book might be wrong. What do you say about it?

Homework Helper
Gold Member
By removing the 26C16 and 6C2 part I am very close to the answer. The numerator of my answer is 18090 whereas that of the answer given in my book is 18080. My denominator is correct. So I think there is a possibility that the answer in my book might be wrong. What do you say about it?
What denominator would that be?

Homework Helper
What denominator would that be?

The answer I get doesn't have a numerator anything like 18080, unless the denominator is a funny looking rational. Are you trying to match multiple choices by making random changes?

Gold Member
What denominator would that be?

Homework Helper

They are both wrong.

Gold Member
They are both wrong.

How?

Homework Helper
How?

Actually, I may not have read your question quite accurately. I read it as exactly one unique letter is repeated twice. As in "exactly two letters of your six letter word are the same". If it means two unique letters of your alphabet are repeated in the six letter word, I'll have to rethink this. This would make your first post make more sense. But I think it might mean the former. It's a little ambiguous. Is "two letters" talking about two letters in the alphabet or two letters in the word?

Last edited:
Homework Helper
Gold Member

I thought of a number of interpretations of the question, but all give me much larger numbers.
A. Of the distinct letters in the 'word', exactly two occur more than once. So the counts of distinct letters could be 4, 2; 3, 3; 3, 2, 1; 2, 2, 1, 1.
B. Exactly two occur twice each, the other two once each, i.e. just the 2,2,1,1 case.
c. As B, but the repeats are immediate, i.e. the two occurrences of a letter are consecutive.

For B, there are 16C2 ways of choosing the two repeated letters, and 14C2 ways of choosing the other two. Having chosen the letters, we must now order them. There are 6C2 places for the first of the repeated letters, 4C2 for the other repeated letter, and two ways of filling in the remaining two digits. 16C2*14C2*6C2*4C2*2 is far larger than either of the answers you quote.

Homework Helper
Dearly Missed

## Homework Statement

A word of 6 letters is formed from a set of 16 different letters of English alphabet(with replacement). Find out the probability that exactly two letters are repeated.

## The Attempt at a Solution

Total possible words = 26C16 . 16^6
Now there are 16C2 ways to choose two letters. Letter 1 can occur atleast twice and atmost 4 times. Same is the case with letter 2. So there arises 5 cases by which repetitions can occur.

Total words possible in these cases are (6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)

∴n(P) = 26C16.16C2.(6C2.4C2.14.13+6C2.4C3.14+6C2.4C4+6C3.3C2.14+6C4.2C2)
P=n(S)/n(P)
But the answer which I get is incorrect.

The way I would do it is to note first that there are C(16,2) = 120 ways of choosing the two letters to be repeated, and for each such way we have the same probability of occurrence; so we might as well assume the repeated letters are A and B, with A repeated a times and B repeated b times. Here, 2 ≤ a,b ≤ 4 and a+b ≤ 6. For the case a=b=2 the remaining two letters are chosen without repetition from 14, and the number of distinct such pairs are C(14,2) = 91, with each such pair having the same probability; so we might as well assume the letters are AABBCD; the probability of this is obtained from a multinomial distribution with 5 categories (A,B,C,D,other) and we want the probability p_5(2,2,1,1,0). So, P{AA,BB} = 91*p_5(2,2,1,1,0). For the case a=3,b=2 there are three classes (A,B,other) and we want p_3(3,2,1), so P(AAA,BB} = p_3(3,2,1). Similarly, P{AA,BBB} = p_3(2,3,1) = P{AAA,BB}. Finally, P{AAAA,BB}=P{AA,BBBB} = p_3(4,2,0) and P{AAA,BBB} = P_3(3,3,0).

Last edited: