How many 4-digit even numbers can be formed using digits 1 to 6 without repetition?

To form a 4-digit even number using digits 1 to 6 without repetition, we need to consider the constraints on the last digit (it must be even) and the fact that all digits used must be distinct. The available digits are 1, 2, 3, 4, 5, 6. The even digits among these are 2, 4, 6 (3 even digits). We can build the 4-digit number by filling the positions from right to left to handle the even number constraint first. Case 1: The last digit is 2. If the last digit is 2, the remaining three digits must be chosen from the remaining 5 digits (1, 3, 4, 5, 6) and arranged in the first three positions. The number of ways to do this is P(5,3)=5×4×3=60. Case 2: The last digit is 4. If the last digit is 4, the remaining three digits must be chosen from the remaining 5 digits (1, 2, 3, 5, 6) and arranged in the first three positions. The number of ways to do this is P(5,3)=5×4×3=60. Case 3: The last digit is 6. If the last digit is 6, the remaining three digits must be chosen from the remaining 5 digits (1, 2, 3, 4, 5) and arranged in the first three positions. The number of ways to do this is P(5,3)=5×4×3=60. The total number of 4-digit even numbers is the sum of the numbers from each case: Total = (Numbers ending in 2) + (Numbers ending in 4) + (Numbers ending in 6) Total = 60+60+60=180. Alternatively, we can think of it in terms of filling the positions: 1.     Choose the last digit (units place): It must be even, so there are 3 choices (2, 4, or 6). 2.     Choose the first digit (thousands place): After choosing the last digit, there are 5 remaining digits. So, there are 5 choices for the first digit. 3.     Choose the second digit (hundreds place): After choosing the first and the last digits, there are 4 remaining digits. So, there are 4 choices for the second digit. 4.     Choose the third digit (tens place): After choosing the first, second, and last digits, there are 3 remaining digits. So, there are 3 choices for the third digit. Total number of even numbers = (Choices for 1st digit) × (Choices for 2nd digit) × (Choices for 3rd digit) × (Choices for last digit) Total number of even numbers = 5×4×3×3=180.