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

2625 namely sum of three options:

All (4) digits are even: 4 possibilities for 1st digit (even but not 0) * 5 for 2nd digit (even) * 5 for 3rd * 5 for 4th = 500

First digit is odd: possibilities are 5 (1st digit odd i.e. in ‘13579’) * 5 (2nd digit even i.e. in ‘02468’) * 5 (3rd digit even) * 5(4th digit even) = 625

Second, third or fourth digit is odd: 4 (1st digit even but not 0) * 5 (even digit) * 5 (even digit) * 5 (odd digit) * 3 (odd digit can be in 2nd, 3rd or 4th position) = 1500

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part