Question: 3-Digit Odd Numbers

Comment on 3-Digit Odd Numbers

Shouldn't the 1st digit be able to completed in 4 ways ?
greenlight-admin's picture

In the solution, we selected the units digit first. This can be accomplished in 2 ways.
My second task was to select a hundreds digit. Since I had already selected ONE digit for the units digit (and since repeated digits are not allowed), there were only 5 digits to choose from (the original 6 digits MINUS the 1 digit I selected to be the units digit)
Does that help?

I am generally confused when to use counting and when to using combination. In this case it seemed like 2* 2C5. How do i distinguish between these two things ?
greenlight-admin's picture

In the MAJORITY of counting questions on the GRE, the Fundamental Counting Principle is all you need. However, there ARE times when you'll need to use some combinations. The following video addresses that very issue:

Fayes question is similar to mine. After choosing ways in which to accomplish the restrictive stage, the remaining unused digits are 4. However u say it's 5 ways to accomplish 1st stage. This becomes confusing at that stage
greenlight-admin's picture

Our first step is to choose a digit (a SINGLE digit) to go in the units place. There are two ways to accomplish this step (we can choose EITHER 1 OR 5)

Once we have placed a SINGLE digit in the units position, there are 5 digits remaining. So, we can complete the next step in 5 ways.

Does that help?


Hi Brent,
Generally, in India these kind of questions are thought with the formulas of permutation and combination. Npr and Ncr
greenlight-admin's picture

Yes, many resources refer to these questions as "Permutation and Combination questions."

I'm not a big fan of this term since it suggests that all counting questions can be solved using either combinations or permutations, when this is not so.

I ramble (on and on) about this in the following article:


Dear Brent - first, kudos you your videos - they are excellent. I need your help on this one please. On a decision tree, my choices are:-
- [1] x [2,4,6,8] x [5], which is 1x4x1= 4 possibilities OR
- [5] x [2,4,6,8] x [1], which is 1x4x1= 4 possibilities OR
- [2,4,6,8 - start with the evens] x [3 remaining even digits left to be chosen for 2nd digi] x [5,1 = the odds] = 4 x 3 x 2 = 24
So I arrive to 24 + 4 + 4 = 32 possibilities.
Thank you.
greenlight-admin's picture

This approach (while much longer than the video solution) will work. The difficult thing is to make sure you've accounted for every possible case that will yield an odd integer.

Your first two cases cover all numbers in the form ODD-EVEN-ODD (total of 8 numbers)
Your third case covers numbers in the form EVEN-EVEN-ODD (24 numbers)

At this point, we're still missing numbers in the form EVEN-ODD-ODD
Number of possible integers = 4 x 2 x 1 = (8 numbers)

So, add all possible cases to get: 8 + 24 + 8 = 40


Thank you.

I am a bit lost on this question I did it the same way as your approached it. but i thought there would be two cases the case for where the units digit is 1 and the second case where the units digit is 5. so i got 40 twice and therefore resulted in 80? why is that approach wrong?
greenlight-admin's picture

The question can be solved by considering two different cases:
1) numbers in the form XY1
2) numbers in the form XY5

For numbers in the form XY1, there are 5 options for the 1st digit and 4 options for the 2nd digit for a total of 20 outcomes.
For numbers in the form XY5, there are 5 options for the 1st digit and 4 options for the 2nd digit for a total of 20 outcomes.

So, we have a total of 40 different outcomes.

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