1. **Problem 14.1.1:** How many different 4-digit numbers can be formed using digits 1, 3, 4, 6, 7, 9 if each digit is used once only? Step 1: We have 6 digits and want to form 4-digit numbers with no repetition. Step 2: The number of such 4-digit numbers is the number of permutations of 6 digits taken 4 at a time: $$P(6,4) = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{720}{2} = 360$$ Answer: 360 different 4-digit numbers. 2. **Problem 14.1.2:** How many 4-digit numbers can be formed that are even and greater than 6000 using digits 1, 3, 4, 6, 7, 9 with no repetition? Step 1: The number must be greater than 6000, so the first digit can be 6 or 7 or 9 (digits greater or equal to 6). Step 2: The number must be even, so the last digit must be an even digit from the set {4, 6}. Step 3: Digits available: 1, 3, 4, 6, 7, 9. Step 4: Consider cases based on the first digit: - Case 1: First digit = 6 - Last digit (even) can be 4 only (since 6 is used as first digit). - Middle two digits chosen from remaining digits {1,3,7,9} without repetition. - Number of ways to choose and arrange middle two digits: $P(4,2) = 4 \times 3 = 12$ - Case 2: First digit = 7 - Last digit (even) can be 4 or 6. - For last digit = 4: - Middle two digits from {1,3,6,9} (excluding 7 and 4) - Number of ways: $P(4,2) = 12$ - For last digit = 6: - Middle two digits from {1,3,4,9} - Number of ways: $P(4,2) = 12$ - Case 3: First digit = 9 - Last digit (even) can be 4 or 6. - For last digit = 4: - Middle two digits from {1,3,6,7} - Number of ways: $P(4,2) = 12$ - For last digit = 6: - Middle two digits from {1,3,4,7} - Number of ways: $P(4,2) = 12$ Step 5: Total numbers = Case 1 + Case 2 + Case 3 $$= 12 + (12 + 12) + (12 + 12) = 12 + 24 + 24 = 60$$ Answer: 60 such numbers. 3. **Problem 14.2.1:** How many different photos can be taken with 5 people standing in a row chosen from 6 doctors, 4 dentists, and 3 nurses with no restrictions? Step 1: Total people available = 6 + 4 + 3 = 13. Step 2: We want to select and arrange 5 people in a row. Step 3: Number of ways = permutations of 13 people taken 5 at a time: $$P(13,5) = \frac{13!}{(13-5)!} = \frac{13!}{8!} = 13 \times 12 \times 11 \times 10 \times 9 = 154440$$ Answer: 154440 different photos. 4. **Problem 14.2.2:** What is the probability that a photo of 5 people will have the 3 nurses sitting next to each other? Step 1: Total number of photos (from 14.2.1) = 154440. Step 2: Treat the 3 nurses as a single block since they must sit together. Step 3: Now, we have this block + other people to fill 5 positions. Step 4: Number of people excluding nurses = 6 doctors + 4 dentists = 10. Step 5: We need to select 2 more people from these 10 to join the block of nurses to make 5 people total. Number of ways to choose 2 people from 10: $$\binom{10}{2} = 45$$ Step 6: Now, arrange the block and the 2 chosen people in a row: Number of ways to arrange 3 entities (block + 2 people): $$3! = 6$$ Step 7: Inside the block, the 3 nurses can be arranged among themselves in: $$3! = 6$$ Step 8: Total favorable arrangements: $$45 \times 6 \times 6 = 1620$$ Step 9: Probability = favorable / total $$\frac{1620}{154440} = \frac{1620}{154440} = \frac{27}{2574} \approx 0.01049$$ Answer: Probability is approximately 0.0105 or about 1.05%.