If you randomly select 6 letters from the word "REGRESSION" and arrange them to form a new meaningless word (where repetition of letters is possible), what is the probability that the new word will have exactly 2 pairs of identical letters?

To solve this problem, we need to calculate the probability that a randomly selected set of 6 letters from the word "REGRESSION" forms a new word with exactly 2 pairs of identical letters. Let's proceed step-by-step:

### Step 1: Analyze the structure of the word "REGRESSION"
The word "REGRESSION" has 10 letters in total, and the distribution of these letters is as follows:

- R: 2 occurrences
- E: 2 occurrences
- G: 1 occurrence
- S: 2 occurrences
- I: 1 occurrence
- O: 1 occurrence
- N: 1 occurrence

So, the unique letters are R, E, G, S, I, O, N, and the frequency distribution of the letters is {R: 2, E: 2, G: 1, S: 2, I: 1, O: 1, N: 1}.

### Step 2: Understand the problem requirements
We are tasked with selecting 6 letters and arranging them such that there are exactly **2 pairs of identical letters**. This means that:
- There are two different letters that each occur twice (these form the two pairs of identical letters),
- The remaining two letters must be distinct and occur exactly once each.

### Step 3: Choose the letters
We need to choose:
1. **Two different letters that each appear in pairs** from the set of letters with frequency 2, which are R, E, and S. The number of ways to choose 2 letters from these 3 is given by:
   \[
   \binom{3}{2} = 3
   \]
2. **Two additional distinct letters** from the remaining pool of letters, which includes all distinct letters {G, I, O, N} as well as the third unused letter from R, E, S. Since we are choosing 2 distinct letters from this set of 5, the number of ways to do so is:
   \[
   \binom{5}{2} = 10
   \]

### Step 4: Arrange the chosen letters
Once we have chosen the letters, the next step is to arrange them. The selected 6 letters will consist of two pairs of identical letters and two distinct letters, so the arrangement is:
\[
\frac{6!}{2!2!1!1!} = \frac{720}{4} = 180
\]
where the \(2!2!\) accounts for the fact that we have two repeated pairs.

### Step 5: Calculate the total possible arrangements
The total number of ways to choose 6 letters from the 10 letters of "REGRESSION" (with repetition allowed) and arrange them is:
\[
\frac{10!}{(10-6)!} = \frac{10!}{4!} = 151200
\]
This assumes that repetition of letters is allowed from the original distribution of the letters in the word "REGRESSION."

### Step 6: Compute the probability
Now, we can calculate the probability as the ratio of favorable outcomes (those with exactly 2 pairs of identical letters) to the total possible outcomes:
\[
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3 \times 10 \times 180}{151200} = \frac{5400}{151200} = \frac{1}{28}
\]

Thus, the probability that the new word will have exactly 2 pairs of identical letters is \( \frac{1}{28} \).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the probability of forming a word with exactly one pair of identical letters?
2. How would the probability change if the word had a different letter distribution?
3. Can you calculate the probability of having no identical letters in the new word?
4. How many different arrangements can you form if all letters must be distinct?
5. What if the word had 12 letters—how would the probability be affected?

**Tip**: When calculating probabilities with arrangements, always check if repetitions or restrictions apply to your selections.

If you randomly select 6 letters from the word 'REGRESSION' and arrange them to form a new meaningless word (where repetition of letters is possible), what is the probability that the new word will have exactly 2 pairs of identical letters?

Explore how to calculate the probability of selecting 6 letters from 'REGRESSION' and forming a word with exactly 2 pairs of identical letters. This problem involves combinatorics, probability, and permutations with repetition. Suitable for students in Grades 11-12.

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