Find the number of 4 digit numbers that contain at least 3 even digits

Many times when attempting to count a large number of items, patterns are recognized and shortcuts to count are developed so that one doesn't need to physically count all of the items. Perhaps the trickiest part of counting with these shortcuts is to ensure that one does not over-count (counting a scenario more than once), although it is equally important that we don't under-count (fail to count a situation that applies).

Part 2. Special circumstances

It should be noted that for a number to be a 4 digit number, it must have 4 digits, and thus the first digit must not be zero (despite that that would be an even digit, the number itself would not be a 4 digit number).

Part 3. Partitioning the situation

There are 2 main scenarios:

1. All 4 digits are even

2. Exactly 3 digits are even (meaning that exactly one digit is odd).

There are two sub-cases to Scenario 2:

2.1. The first digit is odd, and all three of the other digits are even.

2.2. The first digit is even, and exactly two of the other digits are even (meaning that exactly one of the last three digits is odd).

None of scenarios 1, 2.1, and 2.2 overlap, so we're not over-counting:

If all 4 digits are even, then there can't be exactly 3 even digits.
If the first digit is odd, then not all 4 digits are even nor is the first digit even.
If exactly two of the last three digits are even, then not all 4 digits are even, nor are all three of the last digits even.

Further, this is all of the possibilities for a 4 digit number with least three even digits, so we're not under-counting.

Scenario 1 -- All 4 digits are even.

If all 4 digits are even, then the first digit has fours choices (2,4,6,8), and the next 3 digits each have 5 choices (2,4,6,8,0).

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.1 -- The first digit is odd, and all three of the other digits are even

If the first digit is odd, there are 5 choices for the first digit (1,3,5,7,9), and the next 3 digits each have 5 choices (2,4,6,8,0).

[tex]5*5*5*5=625\text{ choices}[/tex]

Scenario 2.2 -- The first digit is even, and exactly two of the other digits are even

If the first digit is even, there are 4 choices for the first digit (2,4,6,8), and if exactly two of the next 3 digits are even, then there are 5 choices for each of the two even digits (2,4,6,8,0), and 5 choices for the odd digit (1,3,5,7,9), and there are "3 permuted by 2" ways of ordering those three digits.

[tex]4* \left [(5*5*5) * {}_3 \! P_2 \right ]=\\\\=4*\left [5*5*5 * \dfrac{3!}{2!} \right ]\\\\=4*\left [ 5*5*5 * \dfrac{3*2*1}{2*1} \right ] \\\\=4*\left [5*5*5 * 3 \right ] \\\\=1500\text{ choices}[/tex]

Scenario 2.2 broken down (calculated without using the permutation operation) -- The first digit is even, and exactly two of the other digits are even

Then either the second digit is odd (scenario 2.2.1), the third digit is odd (scenario 2.2.2), or the fourth digit is odd (scenario 2.2.3).

Scenario 2.2.1. The second digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second digit is odd: 5 choices for the odd digit (1,3,5,7,9)

The third and fourth digits are even: 5 choices for each even digit (2,4,6,8,0).

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.2.2. The third digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second and fourth digits are even: 5 choices for each even digit (2,4,6,8,0).

The third digit is odd: 5 choices for the odd digit (1,3,5,7,9)

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.2.3. The last digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second and third digits are even: 5 choices for each even digit (2,4,6,8,0).

The last digit is odd: 5 choices for the odd digit (1,3,5,7,9)

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenarios 2.2.1, 2.2.2, and 2.2.3 comprise all of Scenario 2.2, so [tex]500+500+500=1500\text{ choices}[/tex]

Part 4. Conclusion

How many ways can a 4 digit number be formed where at least 4 of the digits are even?

This is the sum of the choices from Scenario 1, Scenario 2.1, and Scenario 2.2, so [tex]500+625+1500=2625\text{ choices}[/tex].