How many even numbers of at least four digits can be formed using the digits 0, 1, 2, 3, and 5 without repetition?

Question

I know that the answer 84, but I just can't seem to get that number

Answer 1

 

4 digit numbers:

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    ends with zero, there are 4 choices for the first digit, 3 choices for the 2nd digit, and 2 choices for the 3rd digit.

                4x3x2 = 24

 

    ends with 2:  there are only 3 choices for the first digit because the first digit cannot be zero, as that

                         would make it 3 digit number.  there are 3 choices for the 2nd digit, and 2 choices for the 3rd digit.

                               3 x 3 x 2 = 18.

 

  So there are 24 + 18 = 42 ways to create an 4 digit even integer.

 

5 digit number:

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ending with zero:  there are 4 choices for the first digit, 3 choices for the 2nd digit, 2 choices for the 3rd digit, and

                            one digit remaining for the 4th digit.  4x3x2x1 = 24.

 

ending with 2:   There are only 3 choices for the 1st digit because zero in the first digit makes it a 4 digit number.

                        There are 3 choices for the 2nd digit, 2 choices for the 3rd digit, and one digit remaining for

                        the 4th digit. 3 x 3 x 2 x1 = 18.

 

Again there are 24 + 18 = 42 ways to create an even 5 digit integer.

 

42 + 42 = 84