Out of a collection of 7 consonants and 4 vowels, in how many distinct ways can we form a word consisting of exactly 3 consonants and 2 vowels, assuming that order of letters in the word matters?

Introduction / Context:
This problem tests the ability to combine combinations and permutations in one calculation. We first choose which letters will be used and then arrange them in all possible orders. These multi step counting problems are very common in competitive exams and are an essential part of permutations and combinations practice.

Given Data / Assumptions:

• There are 7 distinct consonants.
• There are 4 distinct vowels.
• We need to form a word of length 5 with exactly 3 consonants and 2 vowels.
• No letter is repeated within a single word.
• Order of letters in the word matters, so different arrangements of the same letters count as different words.

Concept / Approach:
The process naturally splits into two stages. First, select which 3 consonants from the 7 and which 2 vowels from the 4 will appear in the word. This selection uses combinations. Second, arrange the chosen 5 letters in all possible orders, which uses permutations. The total number of words is the product of the number of ways of selecting letters and the number of ways of arranging them.

Step-by-Step Solution:
Step 1: Choose 3 consonants out of 7. This can be done in 7C3 ways.Step 2: Compute 7C3 = (7 × 6 × 5) / (3 × 2 × 1) = 35.Step 3: Choose 2 vowels out of 4. This can be done in 4C2 ways.Step 4: Compute 4C2 = (4 × 3) / (2 × 1) = 6.Step 5: After selection, we have 3 + 2 = 5 distinct letters. The number of ways to arrange 5 distinct letters in order is 5! = 120.Step 6: Multiply all factors: total words = 7C3 × 4C2 × 5! = 35 × 6 × 120.Step 7: First multiply 35 × 6 = 210, then multiply 210 × 120 = 25200.
Verification / Alternative check:
A quick sanity check can be done by noting that 25200 is a reasonable large number for forming ordered words from 11 letters under a moderate restriction. If we had used only permutations directly without separating the selection step, we would risk double counting or miscounting. The combination then permutation method is standard and robust.

Why Other Options Are Wrong:
The other numerical options do not result from any correct combination of 7C3, 4C2 and 5!. For instance, 25000 and 25225 are very close to the correct answer but do not equal 35 × 6 × 120. They are likely included as traps for arithmetic mistakes. Only 25200 exactly matches the required calculation.

Common Pitfalls:
Common errors include forgetting to multiply by 5! after selecting the letters, or choosing the wrong combination counts such as 7P3 instead of 7C3. Another mistake is attempting to place consonants and vowels in specific positions prematurely, which is unnecessary here. Following the clear two stage plan avoids such mistakes.

Final Answer:
The number of different words that can be formed is 25200.