Question

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

Answer 1

Answer:

2625

Step-by-step explanation:

First let's see all the possible combinations.

(Even=E, Odd= O)

1) EEEO

2) EEOE

3) EOEE

4) OEEE

5) EEEE

Now let's see what E and O possibly could be

E = 0, 2, 4, 5, 6 and 8 (5)

O = 1, 3, 5, 7, 9 (5)

Now we are just simply gonna multiply

1) EEEO = 4*5*5*5

2) EEOE = 4*5*5*5

3) EOEE = 4*5*5*5

4) OEEE = 5*5*5*5

5) EEEE = 4*5*5*5

The even contains a 0, so you can't put 0 in, so (5-1=4), there are only 4 digits for the even.

500 times 4 = 2000

5⁴ = 625

2000+625= 2625

Answer 2

Answer:

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