[superbrain]

How many three-digits number can be formed from the digits 0,1,2,3,4,5 and 6 if each digits can be used only once?how many of these are odd numbers? How many are greater than 330?

[Unknown008]

I'm assuming that numbers like 012 is not accepted?

If that is so, then you can do it like this:

_ _ _

In the first spot, you can have any of 1, 2, 3, 4, 5 or 6, which means 6 possibilities.

Let's say you pick 1.

1 _ _

In the second spot, you can have any of 0, 2, 3, 4, 5 or 6, which means yet another 6 possibilities. Assume you pick 4.

1 4 _

In the last spot, you can have any of 0, 1, 2, 3, 5 or 6 which means yet another 5 possibilities.

So, all in all, you have, (6 x 6 x 5) possibilities of having a 3 digit number from the set of given digits.

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Find the number that ends with 1, 3 or 5.

So, find all the numbers which end with 1:

_ _ 1

First spot has 5 possibilities, second spot has again 5 possibilities for a total of (5 x 5) = 25 numbers.

Do the same for 3 and 5 and add them all together to get the required answer.

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Those which are greater that 330 start with either 3, 4, 5 or 6, and in the case they start with 3, the next spot should be either 4, 5 or 6. Can you try this out?