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?

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?