How many even numbers of 3 digits can be formed with the digits 1 2 3 4 5 6 digits can be repeated?

Problem 2: How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated? Solution: Answer: 108.

How many 3-digit even numbers can be formed using the digits 0 1 2 3 4 5 if digits are not repeated?

Count the number of 3-digit strings whose last digit is even. This gives 60−8=52 3-digit even numbers using digits from {0,1,2,3,4,5} without repetition.

How many ways can you make a three-digit even number?

So none of the numbers can end in 1, 3, or 5. So there are 3 ways to choose the last (rightmost) digit, 5 − 3 = 2 ways to choose the first (leftmost) digit, an 1 way to pick the middle digit. Answer is 2 × 1 × 3 = 6 ways to make a three-digit even number with no repetitions of the digits.

How many ways can you choose the last digit of 3-digit numbers?

Want 3-digit even numbers, the digits are 1, 2, 3, 4 and 5, and repetitions are not allowed. So none of the numbers can end in 1, 3, or 5. So there are 3 ways to choose the last (rightmost) digit, 5 − 3 = 2 ways to choose the first (leftmost) digit, an 1 way to pick the middle digit.

What is the total number of 3 digit numbers formed?

Now there are 4 numbers left so you can fill the second place in ways and similarly 3rd place in 3 ways. Hence, total number of 3-digit numbers formed = 5*4*3 = 60. Since the numbers have to be even, the digit in the unit place can be either 2 or 4.

Why do we not use all the possible combinations of numbers?

Not using all digits allows more possibilities. (So if 2 is the last digit, and there is no 1 used, there are 3!=6, 3! for the swap = 12 more. 12 more for no 3, 12 more for no 5. If there is no 4 then just 6 ways, same 6 more if there is no 2. +48 more ways). Not using 2 digits is left to the reader.