Hi.. can somebody explain the following problem... please, :I

.) how many 3 digit natural no's can be formed from the digits 0,1,2,3,4,5,6 if repetition is not allowed
1) if repetition is allowed
2)how many of them are odd and even numbers?

ANY HELP WILL BE USEFUL AND APPRECIATED, THANKS

What is your attempt?

Ive never faced such questions in this chapter.. :-(

Okay, here's an example:

From the digits 0, 1, 2, 3, for the number of 3 digit numbers (without repetition) that can be formed can be worked out this way:

3 digit numbers: 3 'slots'.

_ _ _

The first slot can contain either of 1, 2, 3 or 4. Hence 4 possibilities. (we cannot have 0 in front since that would make it a 2 digit number)

4 _ _

Say 3 was taken. There are now 0, 1, 2, 4 left. The second slot can contain any of those, for a total of 4 possibilities.

4 4 _

Say 0 is taken now. What's left is 1, 2, 4: which mean 3 possibilities.

4 4 3

Total number of 3 digit numbers = 4 * 4 * 3 = 48 numbers

With repetition, you'd have:
4 5 5

That is: 4 * 5 * 5 = 100 numbers

Can you see why?

Now, can you apply this to your problem?

Yes.. its helpful.. But having problem in even odd questions...

Then, fill the last digit first. What digits can go there if the number is to be even, or odd?

3*3*2=18 for odd..

For your problem, you are required to make 3 digit numbers from the digits 0,1,2,3,4,5,6

_ _ _

The last slot can be 1, 3 or 5, hence only 3 possibilities.

_ _ 3

Since 333 is allowed (repetition), the second number can be any of 0,1,2,3,4,5 or 6 (7 possibilities).

_ 7 3

The first can be any of: 1,2,3,4,5 or 6 (6 possibilities)
6 7 3

6*7*3 = 146 different odd numbers

Thanks... :)