The letters from the word PERIMETER are arranged in a row.

1. Identify the total number of letters and the frequencies of any repeating letters

The word PERIMETER consists of 9 letters.
The letter 'E' appears 3 times and the letter 'R' appears 2 times.

2. Calculate the total number of arrangements by dividing the factorial of the total letters by the factorials of the repeating letters

$\text{Number of arrangements} = \frac{9!}{3! \times 2!} = 30240$

1. Determine the possible pairs of positions for the two identical R's.

For there to be at least 6 letters between the two R's in a 9-letter word, the two R's can only be placed in the following position pairs:

• 1st and 8th (6 letters between)
• 1st and 9th (7 letters between)
• 2nd and 9th (6 letters between)

There are 3 possible ways to position the two R's.

2. Calculate the number of arrangements for the remaining letters.

The remaining 7 letters are P, E, I, M, E, T, E.
Since there are three identical E's, the number of ways to arrange these 7 letters in the remaining 7 spaces is:
$\frac{7!}{3!} = \frac{5040}{6} = 840$

3. Calculate the total number of arrangements.

Total number of arrangements $= 3 \times 840 = 2520$

1. Find the total number of possible arrangements of the remaining 8 letters.

After removing one 'E', the 8 remaining letters are P, E, E, R, R, I, M, T.
The total number of unrestricted arrangements is:
$n(S) = \frac{8!}{2!2!} = 10080$

2. Find the number of arrangements where the two E's are together and where the two R's are together.

To find arrangements with the two E's together, treat 'EE' as a single block. We then arrange 7 items (EE, R, R, P, I, M, T):
$n(\text{EE}) = \frac{7!}{2!} = 2520$

Similarly, to find arrangements with the two R's together, treat 'RR' as a single block. We arrange 7 items (E, E, RR, P, I, M, T):
$n(\text{RR}) = \frac{7!}{2!} = 2520$

3. Find the number of arrangements where both the E's and the R's are together.

Treat both 'EE' and 'RR' as single blocks. We then arrange 6 items (EE, RR, P, I, M, T):
$n(\text{EE and RR}) = 6! = 720$

4. Apply the principle of inclusion-exclusion to find the valid arrangements, then compute the probability.

Number of arrangements where at least one pair is together:
$n(\text{EE or RR}) = n(\text{EE}) + n(\text{RR}) - n(\text{EE and RR})$
$= 2520 + 2520 - 720 = 4320$

Number of arrangements with no identical adjacent letters:
$10080 - 4320 = 5760$

$\text{Probability} = \frac{5760}{10080} = \frac{4}{7}$