How many natural 4-digit numbers exist in which all of the digits are even?

500

Step-by-step explanation:

So first off we know that there are 9000 (9k) natural 4-digit numbers -->

we solve using permutations so we can set the problem up like this

1          2          3           4

4     x     5     x     5     x      5  = 100 x 5 = 500

Reasoning:

So the reason we can do it like this is because -->

even numbers are: 0, 2, 4, 6, 8

ALL DIGITS MUST BE EVEN --> calculate using permutations simply makes it 5^4 but that's NOT our answer. Since the first digit can't be zero, we have -->

4, 5, 5, 5 respectively

Multiply this (by rule of permutations) to get, => 20 x 5 = 100

100 x 5 = 500

Thus, we have our answer of 500

Hope this helps!

c is the answer

step-by-step explanation:

the answer is c just got it on ed

step-by-step explanation: