From a group of 5 consonants and 4 vowels, in how many different words can be formed using exactly 3 consonants and 2 vowels, assuming that order of letters in the word matters?

Introduction / Context:
This question checks your understanding of combinations and permutations together. We are forming words (ordered arrangements of letters) from a specified pool of consonants and vowels, with the constraint that each selected word must contain exactly 3 consonants and 2 vowels. The problem explicitly involves choosing which letters to use and then arranging them, so both selection and ordering steps must be handled carefully.

Given Data / Assumptions:
- Available consonants: 5 distinct consonants.
- Available vowels: 4 distinct vowels.
- Each word must have exactly 5 letters in total.
- Composition of each word: 3 consonants and 2 vowels.
- No letter is repeated in a single word.
- Order of letters in a word is important, so different permutations count as different words.

Concept / Approach:
The process naturally splits into two stages. First, choose which consonants and vowels will appear in the word. Second, arrange the chosen 5 letters in all possible orders. The number of ways to choose consonants uses combinations from the 5 available consonants, and likewise for vowels. After choosing 3 consonants and 2 vowels, we have 5 distinct letters, and these can be arranged in 5! different ways. The final count is the product of counts from these two main stages.

Step-by-Step Solution:
Step 1: Choose 3 consonants out of 5 available consonants.Step 2: Number of ways to do this is 5C3 = 10.Step 3: Choose 2 vowels out of 4 available vowels.Step 4: Number of ways to do this is 4C2 = 6.Step 5: After selections, we have 5 distinct letters (3 consonants + 2 vowels).Step 6: Arrange these 5 letters in all possible orders. The number of permutations is 5! = 120.Step 7: Multiply the three factors: 5C3 * 4C2 * 5! = 10 * 6 * 120.Step 8: Compute: 10 * 6 = 60; 60 * 120 = 7200.

Verification / Alternative check:
An alternate viewpoint is to select positions first for consonants or vowels and then fill them. However, because the letters themselves are distinct, this re-phrasing still leads to the same product 10 * 6 * 120. Using a small toy example with fewer letters helps demonstrate that each step corresponds to an independent choice, and the basic rule of product (multiplication principle) justifies multiplying the counts from each stage.

Why Other Options Are Wrong:
- 7600 and 6400 usually arise from arithmetic errors, such as miscomputing combinations or mis-evaluating 5! as 100 or 130 instead of 120.
- 3600 might come from mistakenly dividing the total by 2 or halving at some step, perhaps wrongly assuming that order does not matter fully.

Common Pitfalls:
Some learners forget to separate the selection of letters from the arrangement and attempt to count everything directly as permutations, which can lead to confusion or double-counting. Others mistakenly treat combinations as if order mattered and use permutations in the selection phase. Maintaining a clear distinction—combinations for choosing which letters, permutations for ordering them—helps keep the logic straight and yields the correct result.

Final Answer:
The number of different words that can be formed using exactly 3 consonants and 2 vowels is 7200.