Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.
- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos
- Study Guide
- Philosophy
- Office Hours
- Extras
- Prices
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
I am a bit lost on this
The question can be solved by
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.