Calculation: Total digit = 6 {0, 1, 2, 3, 4 and 5} We have to formed 3-digit even numbers: Case-1: When 0 is at unit place, then Number of choices for the units place = 1 (0 is at unit place) Number of choices for the ten's place = 5 (Any digit except the one that is already used as the 1st digit) Number of choices for the hundredth's place = 4 (Any digit except the ones that are already used as the 1st and 2nd digit) No. of such three digit even number = 1 × 5 × 4 = 20 Case-2: When 0 is at ten's place, then For even number unit place must be filled with 2 or 4 Number of choices for the units place = 2 Number of choices for the ten's place = 1 Number of choices for the hundredth place = 4 No. of such three digit even number = 2 × 1 × 4 = 8 Case-3: When no digit in the 3-digit even number is zero. For even number unit place must be filled with 2 or 4 Number of choices for the units place = 2 Number of choices for the ten's place = 4 Number of choices for the hundredth place = 3 No. of such three digit even number = 2 × 4 × 3 = 24 Total number of possible three digit even number = 8 + 24 + 20 = 52 So for 100th place we have 9 digits, for tenth place we have 10 digits and for units place we have 10 digits So total 3 digits number are 9*10*10 =900 If we talk about even numbers then we have only 5 digits fot units place 0,2,4,6,8 So total 3 digits even numbers would be 9*10*5 =450 How many 3If z is 0, then x has 9 choices. Therefore, z and x can be chosen in (1 × 9) + (4 × 8) = 41 ways. For each of these ways, y can be chosen in 8 ways. Hence, the desired number is 41 × 8 = 328 numbers 3-digit even numbers exist with no repetitions.
How many 3 digits are even numbers?Thus, by multiplication principle, the required number of 3-digit numbers is 3×20=60.
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