hey my name is kevin,
i was wondering how many possibilities is there for a four digit number with the same first two numbers and the last two is not one of the first with numbers 1-9


Hi Kevin.

Think of it this way: The first and second number are the same always, so you could treat them as a single digit if you want. There are nine possible digits (1-9), so there are nine choices for the first pair of digits.

Now you look at the last two. Each can be any digit (1-9) except for the digit used in the first pair. That means 8 choices for each. There is no restriction that they depend on each other at all.

So we have a doublet (nine choices), a "third" digit (eight choices), and a last digit (eight choices). When you have several independent choices and are trying to figure out how many possibilities they represent, you multiply.

This problem is the kind we'd start with when discussing the ideas of "permutations and combinations". You can see another example in a question we received from Ken or search on "permutations" or "combinations" in the Quandaries & Queries search.

Stephen La Rocque>