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## Comment on

3-Digit Odd Numbers## Shouldn't the 1st digit be

## In the solution, we selected

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

## In the MAJORITY of counting

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: https://www.greenlighttestprep.com/module/gre-counting/video/788

## Fayes question is similar to

## Our first step is to choose a

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?

Cheers,

Brent

## Hi brant generally, in India

Generally, in India these kind of questions are thought with the formulas of permutation and combination. Npr and Ncr

## Yes, many resources refer to

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: https://www.gmatprepnow.com/articles/combinations-and-non-combinations-%...

Cheers,

Brent

## Dear Brent - first, kudos you

- [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.

## This approach (while much

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

Cheers,

Brent

## Thank you. Regards

Regards

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