Question

how many different three digit even number can be formed using the digit 0,1,2,3,4,5 if each digit is to be used at most once ? Plz provide solution .... & plz don't spam!!!!

Answer 1

Okay, so let's see this step by step.

As we know even numbers are those integers which have 0 or 2 or 4 or 6 or 8 at the unit's place.

Since we want three digit even numbers with no repetition, we are seeking for numbers which end with 0/2/4/6/8 and do not start with 0.

Case 1: Numbers ending with 0.

Since they already have 0 in the unit's place, some other digit should occupy the 10th's place.

There are 6 other digits which can occupy this place.

Now let's come to 100th's place. Apart from 0 and the digit that's already put in the 10th's place, there are 5 distinct digits which may now occupy the 100th's place.

Thus, total number of combinations = 5 * 6 = 30

Case 2: Numbers ending with 2 or 4 or 6

We now have 3 options to choose from and put at the unit's place.

Let say we choose some digit (say 2) and put it in the unit's place.

Now that we've already used 2, it cannot be used again in the remaining places.

Additionally we've one more condition that we cannot start our number with 0. So now if we consider 100th's place, 0 and 2 have already been used so we're left with 5 options for 100th's place.

After filling the 100th's place, let's come to 10th's place.Here we cannot use 2, and we cannot use the number we already used at 100th's place. But wait, here we can use 0, right? So again we've 5 options for this place.

Thus total number of combinations ending with 2 = 5 * 5 = 25

(The same calculation would follow for numbers ending with 4 and 6 as well, so that would contribute 25 combinations each.)

Thus,

Total number of even combinations without repetition of digits

= 30 + (3*25) = 105

Hope this helps :)

peace ✌