The problem asks for the probability that a random permutation of the letters in the word "PEPPER" yields the same word ("PEPPER"). The answer should be an irreducible fraction p/7 with an integer p.

1. Problem Description

The problem asks for the probability that a random permutation of the letters in the word "PEPPER" yields the same word ("PEPPER"). The answer should be an irreducible fraction p/7 with an integer p.

2. Solution Steps

First, we need to find the total number of possible permutations of the letters in the word "PEPPER". The word has 6 letters, with "P" appearing 3 times and "E" appearing 2 times, and "R" appearing once. The formula for permutations with repetitions is:

$N = \frac{n!}{n_1! n_2! ... n_k!}$

where $n$ is the total number of items, and $n_i$ is the number of repetitions of the $i$-th item.

In our case, $n = 6$, $n_1 = 3$ (for "P"), $n_2 = 2$ (for "E"), and $n_3 = 1$ (for "R").

So the total number of permutations is:

$N = \frac{6!}{3! 2! 1!} = \frac{720}{(6)(2)(1)} = \frac{720}{12} = 60$

There is only 1 arrangement that spells "PEPPER."

The probability of a random permutation resulting in "PEPPER" is:

$P = \frac{1}{N} = \frac{1}{60}$

The question requires the answer to be in the form $p/7$, where p is an integer. Since $1/60$ needs to equal $p/7$, it seems that the problem statement has a typo, because we are not able to make $1/60$ into a fraction with denominator 7 with an integer numerator. If instead we need the answer to be $p/n$ where $n$ is as close as possible to

7. Then consider the following:

$\frac{1}{60} \approx \frac{x}{7}$

$x = \frac{7}{60} \approx 0.11666$

If we multiply by $60/7$, $60/7 \times (1/60) = 1/7$.

Since the irreducible fraction is $1/60$ we want it to be $p/7$ and p an integer. This is impossible.

However, let's assume the problem wanted to write the probability as an approximate fraction with 7 in the denominator and we are looking for an integer as close as possible. The best approximation will be $1/60 \approx p/7$ which we can solve:

$p=7/60 = 0.116$. Since $p$ is meant to be an integer we can take $0$. $0/7 = 0$.

Let us check if there is a possible typo in the problem. Perhaps the total arrangements of letters are to sum up to $x \times 7$ for some integer $x$. $60 = 8 \times 7 + 4$

Consider if instead we write $N = 56$ which is divisible by

7. Then total permutations becomes $56/60 = 14/15$

Alternatively maybe they want to write $1/60 = p/q$ and we need to find the fraction closest to this value such that $q$ is $7$.

3. Final Answer

Given the constraints, the original question is misleading. However, assuming the problem intended an approximation and an integer p, then p =

0. So the answer is

0.

Let us assume there's a typo in the prompt. Assume the question is "express the result as an irreducible fraction."

Then the answer is 1/

6

0. Since we must give p/7, we can't directly.

Let's assume the problem asked for an approximation. $1/60 = 0.016666$.

$x/7 = 0.01666$, so $x = 0.1166$. The closest integer to $0.1166$ is

0.

Therefore, $p=0$.

Final Answer: 0