How many 3-digit numbers exist, whose digits are distinct even numbers.

48

Step-by-step explanation:

To solve this question will demand use of permutation :

Now, we have 5 single digit even numbers which are 0,2,4,6,8

Now, the hundredth place will be able to hold any digit except 0 because if it is done otherwise, it will make it a two digit number.

Also, the tens digit can now hold four digits as one digit is already placed in hundredth place.

Likewise, the unit digit can now hold any of the remaining three digits.

From the fundamental principle of counting, the required number of three digit numbers is :

4 x 4 x 3 = 48

i need the options

step-by-step explanation:

i think it is 18

step-by-step explanation: