1. **Problem statement:** We want to find how many different codes can be created with 2 letters, 2 digits, and 2 letters in that order, with no repeated letters or digits, and exactly 1 vowel in the entire code. Letters are chosen from the set {A, B, C, D, E, F, G, H, I, J} (10 letters), digits from 0 to 9. 2. **Step I: Number of vowel choices for the single vowel letter.** The vowels in the set are A, E, I (3 vowels). Since there must be exactly 1 vowel in the code, the number of possibilities to choose this vowel letter is: $$3$$ 3. **Step II: Number of ways to choose the 3 consonants.** There are 10 letters total, 3 vowels, so 7 consonants remain. We need to choose 3 consonants without repetition. The number of ways to choose and arrange 3 consonants in the 3 remaining letter positions is the number of permutations of 7 letters taken 3 at a time: $$P(7,3) = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$$ 4. **Step III: Number of ways to choose the 2 digits.** Digits are from 0 to 9 (10 digits), no repetition. Number of ways to choose and arrange 2 digits is: $$P(10,2) = \frac{10!}{8!} = 10 \times 9 = 90$$ 5. **Step IV: Total number of codes with exactly 1 vowel and no repetitions.** The code format is 2 letters, 2 digits, 2 letters. The vowel can be in any of the 4 letter positions (positions 1, 2, 5, or 6). Number of ways to place the single vowel in one of these 4 positions: $$4$$ For each vowel position, the other 3 letter positions are consonants (210 ways), and digits are 90 ways. Total codes: $$4 \times 3 \times 210 \times 90 = 226800$$ **Final answers:** I. 3 II. 210 III. 90 IV. 226800