Experiment: Find all the possible two digit numbers that only use digits 1-9 and no digit repeats.
Event1: choose a digit for your tens place
Event2: choose a digit for your ones place

Number of possible outcomes for Event1: 9
Namely: 1,2,3,4,5,6,7,8,9

Number of possible of outcomes for Event2: 8
Namely: we can choose any number from 1 to 9 except the one chosen in Event1

By Fundamental Counting Principle: # of outcomes in the entire experiment is the product of the number of possible outcomes in both events.
9*8=72


But why can we use Fundamental Counting Principle?
B/C no matter what number we choose in Event1 we'll always have 8 possible outcomes in Event2. That is, the NUMBER of possible outcomes of each event is independent of one another.
If the number of possible outcomes between both events was not independent then we wouldn't be able to use Fundamental Counting Principle.

An alternative method would be to use the process of elimination. There are 100 numbers between 1 and 100 inclusive.

1. Remove 100 and the nine single digit numbers. We are down to 90 numbers.

2. Next, remove the nine two-digit numbers which have the same tens and ones digit. We are down to 81 numbers.

3. Finally, remove the nine two digit numbers ending in zero. We are down to 72 numbers, and thus the answer is B.

We could directly use the permutation formula \(9P2 = \frac{9!}{(9-2)!} = \frac{9!}{7!} = 9 * 8 = 72 \)

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.