1. **Problem statement:** From the digits 1, 2, 2, 3, 4, 5, form 4-digit numbers where each digit is used at most once. Conditions: - The first digit cannot be 2. - The number must contain at least one 2. 2. **Available digits:** 1, 2, 2, 3, 4, 5 3. **Step 1: Identify possible first digits.** - First digit cannot be 2, so possible first digits are 1, 3, 4, 5. 4. **Step 2: The number must contain at least one 2.** - Since the first digit is not 2, at least one of the remaining three digits must be 2. 5. **Step 3: Count total 4-digit numbers with first digit in {1,3,4,5} and at least one 2 in the last three digits.** 6. **Step 4: Calculate total 4-digit numbers with first digit in {1,3,4,5} without restriction on 2's:** - First digit: 4 choices (1,3,4,5) - Remaining digits: choose 3 digits from the remaining 5 digits (including two 2's) without repetition. Number of ways to choose and arrange last 3 digits: - Total digits left after choosing first digit: 5 digits - Number of permutations of 3 digits from 5 digits with duplicates (two 2's) considered: We consider cases based on how many 2's are used in the last 3 digits. 7. **Step 5: Break down by number of 2's in last 3 digits:** - Case 0: No 2's in last 3 digits - Case 1: Exactly one 2 in last 3 digits - Case 2: Exactly two 2's in last 3 digits 8. **Step 6: Calculate Case 0 (no 2's in last 3 digits):** - Remaining digits excluding 2's: {1,3,4,5} minus the first digit - Since first digit is from {1,3,4,5}, exclude it from the set - So, for each first digit, the remaining digits without 2's are 3 digits - Number of ways to arrange 3 digits from these 3 digits is 3! = 6 Number of first digits: 4 Number of Case 0 numbers: 4 * 6 = 24 9. **Step 7: Calculate total 4-digit numbers with first digit in {1,3,4,5}:** - Total digits left after choosing first digit: 5 digits - Number of permutations of 3 digits from 5 digits with duplicates (two 2's) considered: Total permutations without restriction: - Number of ways to select and arrange 3 digits from 5 digits with duplicates: We calculate total permutations: - Total permutations of 3 digits from 5 digits with two identical 2's: Number of permutations = permutations with 0, 1, or 2 twos: - 0 twos: choose 3 digits from {1,3,4,5} (4 digits), permutations = P(4,3) = 4*3*2=24 - 1 two: choose 2 digits from {1,3,4,5} and 1 two, permutations = C(3,1)*P(4,2)* arrangements Actually, better to count permutations directly: Number of permutations with exactly one 2: - Fix one 2, choose 2 digits from 4 digits {1,3,4,5} without repetition: C(4,2)=6 - Number of ways to arrange these 3 digits (1 two + 2 others): 3! = 6 - Total: 6 * 6 = 36 - 2 twos: choose 1 digit from {1,3,4,5} and 2 twos - Number of ways to choose 1 digit: 4 - Number of ways to arrange these 3 digits (2 twos + 1 other): permutations = 3!/2! = 3 - Total: 4 * 3 = 12 Total permutations for last 3 digits = 24 + 36 + 12 = 72 10. **Step 8: Total numbers with first digit in {1,3,4,5} and any last 3 digits =** - 4 (first digit choices) * 72 (last 3 digits permutations) = 288 11. **Step 9: Numbers with no 2 in last 3 digits = 24 (from Step 6)** 12. **Step 10: Numbers with at least one 2 in last 3 digits = Total - no 2's = 288 - 24 = 264** **Final answer:** $$\boxed{264}$$