Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects, without regard to order, is given by the binomial coefficient: ${n \choose k} = \frac{n!}{k!(n-k)!}$
• Fundamental Principle of Counting: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.

Step-by-Step Solution

Step 1: Understand the Problem

We need to select 5 letters from the English alphabet such that they are in alphabetical order, and the middle letter is 'M'. This means we need to choose two letters before 'M' and two letters after 'M'.

Step 2: Choosing Letters Before 'M'

The letters before 'M' are A, B, C, ..., L. There are 12 letters in total. We need to choose 2 of these letters. Since the order is already determined (alphabetical), we use combinations. The number of ways to choose 2 letters from these 12 is: ${12 \choose 2} = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66$

Step 3: Choosing Letters After 'M'

The letters after 'M' are N, O, P, ..., Z. There are 13 letters in total. We need to choose 2 of these letters. Again, since the order is already determined (alphabetical), we use combinations. The number of ways to choose 2 letters from these 13 is: ${13 \choose 2} = \frac{13!}{2!(13-2)!} = \frac{13!}{2!11!} = \frac{13 \times 12}{2 \times 1} = 78$

Step 4: Calculate the Total Number of Ways

Since choosing the letters before 'M' and choosing the letters after 'M' are independent events, we multiply the number of ways for each event to find the total number of ways: ${12 \choose 2} \times {13 \choose 2} = 66 \times 78 = 5148$

Step 5: This result does not match the answer given. We must re-examine our calculations and the problem statement.

The problem states the middle letter must be 'M'. The chosen 5 letters must be in alphabetical order. Let the five letters be $L_1, L_2, L_3, L_4, L_5$. We are given $L_3 = M$. The letters $L_1$ and $L_2$ must be chosen from the 12 letters before M (A to L). The letters $L_4$ and $L_5$ must be chosen from the 13 letters after M (N to Z). Thus we have ${12 \choose 2}$ ways to choose $L_1$ and $L_2$, and ${13 \choose 2}$ ways to choose $L_4$ and $L_5$. The total number of ways is ${12 \choose 2} \times {13 \choose 2} = 66 \times 78 = 5148$.

The correct answer provided, 6084, is incorrect. Let's see if we can derive it.

Suppose we choose the letters with replacement. Then, to choose two letters from the first 12, we have $12 \times 12 = 144$ ways. But since we need them to be in order, we have ${12+2-1 \choose 2} = {13 \choose 2} = 78$, which is still incorrect.

Suppose the correct answer is ${n \choose 5}$. Then ${n \choose 5} = 6084$. We can approximate this as $n^5 / 5! \approx 6084$. Thus $n^5 \approx 6084 \times 120 \approx 730000$. Then $n \approx 15$. This does not make sense.

Let's try to work backwards from the answer choices. We know ${12 \choose 2} \times {13 \choose 2} = 5148$. (A) 6084 (B) 5148 (C) 14950 (D) 4356

Since the correct answer is (A), and we derived 5148, let's see if we made an error. We are choosing 2 letters from 12, and 2 letters from 13. ${12 \choose 2} = \frac{12 \times 11}{2} = 66$ ${13 \choose 2} = \frac{13 \times 12}{2} = 78$ $66 \times 78 = 5148$.

Let's check the other answer choices. (D) $4356 = 66 \times 66 = {12 \choose 2} \times {12 \choose 2}$. (C) $14950 = 2 \times 5^2 \times 598$.

The correct answer provided by the question is incorrect.

Common Mistakes & Tips

• Carefully read the problem statement to understand the constraints.
• Remember the difference between permutations (order matters) and combinations (order doesn't matter).
• Double-check your calculations to avoid arithmetic errors.

Summary

We needed to find the number of ways to choose 5 letters in alphabetical order such that the middle letter is 'M'. We calculated the number of ways to choose two letters before 'M' and two letters after 'M' using combinations and then multiplied these values to get the total number of ways. Our calculation resulted in 5148, but the provided answer is 6084. The derived answer is 5148.

Final Answer The final answer is \boxed{5148}, which does not correspond to any of the options. The question has an incorrect answer.