Question

Find the number of 4-digit numbers that contain at least three even digits.

Answer 1

4625 numbers

Explanation:

To solve this problem we can separate it in different cases, and then we sum the amount of numbers of each case to find our final result.

The first case is: all four numbers are even.

We have 5 even digits, and the first digit of the 4-digit number can't be zero, so for the first digit we have 4 possible values, and for the other 3 digits we have 5 possible values, then the amount of numbers with this condition is:

`N_1 = 4 * 5 * 5 * 5 = 500\ numbers`

The second case is the number having three even digits and one odd digit, and the first digit is the odd digit.

In this case, the first digit has 9 possible values (all digits but zero) and the other 3 digits have 5 possible values (even digits):

`N_2 = 9 * 5 * 5 * 5 = 1125\ numbers`

The third case is the number having three even digits and one odd digit, and the first digit is not the odd digit.

In this case, the first digit has 4 possible values, the odd digit has 10 possible values and the other two digits have 5 possible values. The odd digit can be the second, the third or the fourth digit, so we multiply our result by 3 (because we will find the same result if we choose the odd digit to be the second, third or fourth):

`N_3 = 3 * 4 * 10 * 5 * 5 = 3000\ numbers`

So the final result is the sum of these cases:

`N = N_1 + N_2 + N_3`

`N = 500 + 1125 + 3000 = 4625\ numbers`