Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, where order doesn't matter, is given by: ${n \choose r} = \frac{n!}{r!(n-r)!}$
• Case-Based Reasoning: Dividing a problem into mutually exclusive and exhaustive cases to systematically count possibilities.
• Frequency Analysis: Determining the number of times each distinct item appears in a given set.

Step-by-Step Solution

Step 1: Analyze the word MATHEMATICS and identify distinct letters and their frequencies.

• What we are doing: We are identifying the distinct letters in the word MATHEMATICS and counting how many times each letter appears. This is crucial for determining the constraints on our selections.
• Math & Reasoning:
  • M: 2
  • A: 2
  • T: 2
  • H: 1
  • E: 1
  • I: 1
  • C: 1
  • S: 1 Therefore, we have 3 letters (M, A, T) that appear twice each and 5 letters (H, E, I, C, S) that appear once each. The total number of distinct letters is 8.

Step 2: Define the possible cases based on repetition.

• What we are doing: We are defining the possible scenarios for selecting 5 letters, considering that some letters can be repeated up to a maximum of two times (due to their frequency in the original word).
• Reasoning: We need to choose 5 letters. The possibilities for repetition are:
  • Case I: All 5 letters are distinct.
  • Case II: One pair of identical letters and 3 other distinct letters.
  • Case III: Two pairs of identical letters and 1 other distinct letter. These are the only possible cases because no letter appears more than twice, and we are choosing only 5 letters.

Step 3: Calculate the number of ways for Case I (All 5 letters are distinct).

• What we are doing: We are calculating the number of ways to choose 5 distinct letters from the 8 distinct letters available (M, A, T, H, E, I, C, S).
• Math & Reasoning: We need to choose 5 letters from 8 distinct letters. This is a combination problem. ${8 \choose 5} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ There are 56 ways for Case I.

Step 4: Calculate the number of ways for Case II (One pair of identical letters and 3 other distinct letters).

• What we are doing: We are calculating the number of ways to choose one pair of identical letters (from M, A, or T) and then choose 3 distinct letters from the remaining letters.
• Math & Reasoning:
  1. Choose the letter for the pair: There are 3 options (M, A, T). ${3 \choose 1} = 3$
  2. Choose the 3 distinct letters: After choosing one letter for the pair, we have 7 distinct letters remaining. We need to choose 3 from these 7. ${7 \choose 3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ Total for Case II: ${3 \choose 1} \times {7 \choose 3} = 3 \times 35 = 105$ There are 105 ways for Case II.

Step 5: Calculate the number of ways for Case III (Two pairs of identical letters and 1 other distinct letter).

• What we are doing: We are calculating the number of ways to choose two pairs of identical letters (from M, A, or T) and then choose 1 distinct letter from the remaining letters.
• Math & Reasoning:
  1. Choose the two letters for the pairs: We need to choose 2 letters from the 3 options (M, A, T). ${3 \choose 2} = \frac{3!}{2!1!} = 3$
  2. Choose the 1 distinct letter: After choosing two letters for the pairs, we have 6 distinct letters remaining. We need to choose 1 from these 6. ${6 \choose 1} = 6$ Total for Case III: ${3 \choose 2} \times {6 \choose 1} = 3 \times 6 = 18$ There are 18 ways for Case III.

Step 6: Calculate the total number of ways.

• What we are doing: We are summing the number of ways from each case to find the total number of ways to choose 5 letters from the word MATHEMATICS.
• Math & Reasoning: Total Ways = Case I + Case II + Case III Total Ways = 56 + 105 + 18 = 179

Common Mistakes & Tips

• Be careful to consider the limited repetition allowed. Don't assume unlimited repetition.
• Always analyze the frequencies of the letters first. This helps in defining the cases correctly.
• Ensure the cases are mutually exclusive and exhaustive.

Summary

We have determined the number of ways to choose five alphabets from the word MATHEMATICS, considering the limited repetition of letters, by using case-based reasoning. We analyzed the frequencies of the letters, defined three mutually exclusive and exhaustive cases, calculated the number of ways for each case using combinations, and summed the results to obtain the total number of ways.

The final answer is $\boxed{179}$, which corresponds to option (A).