How many integers between 100 and 999 have all distinct digits (i. e., no two digits the same)?

648




Explanation:

We have this set of digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 
From that set, we can only pick three items. We cannot select the same digit twice.

Consider a blank three digit number such that it is composed of slot A, slot B, slot C. 

Since the number must be larger than 100, this means that we cannot select 0 as the first digit. We go from a pool of 10 digits to 10-1 = 9 digits for our first selection.
In other words, we have this subset to select from
{1, 2, 3, 4, 5, 6, 7, 8, 9}
So we have 9 choices for slot A.

For slot B, we also have 9 choices since 0 is now included. For instance, if we selected the digit '4' then we have this subset of choices left over: {0, 1, 2, 3, 5, 6, 7, 8, 9} which is exactly 9 items. 

For slot C, we have 9-1 = 8 items left to choose from. For example: If we choose '4' for slot A, and '2' for slot B, then we have this subset to choose from: {0, 1, 3, 5, 6, 7, 8, 9} exactly 8 items

In summary so far, we have...
9 choices for slot A
9 choices for slot B
8 choices for slot C

Giving a total of 9*9*8=81*8=648 different three digit numbers. You'll notice that I'm using the counting principle which allows for the multiplication to happen. Think of a probability tree. 

The additive identity is 0 because when you add 0 to a number, you get the identical number as an answer.a + 0 = a