# Difficult GMAT Counting Problems

Here are a few 800+ counting problems.  Yes, that’s right, 800+, which means, as hard as (or possibly harder than) anything you will see on the GMAT.  If you can do these, you are in great shape!

1) How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?

A. 14

B. 24

C. 28

D. 48

E. 96

2) How many distinct four-digit numbers can be formed by the digits {1, 2, 3, 4, 5, 5, 6, 6}?

A. 280

B. 360

C. 486

D. 560

E. 606 3) In the grid of dots above, each dot has equal vertical & horizontal spacing from the others.   A small 45-45-90 triangle is drawn.  Counting this triangle, how many triangles congruent to this one, of any orientation, can be constructed from dots in this grid?

A. 112

B. 120

C. 240

D. 448

E. 480

4) Fifteen dots are evenly spaced on the circumference of a circle.  How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160

B. 450

C. 910

D. 1360

E. 2640

## Basic ideas of counting

The most basic of all ideas in counting is the Fundamental Counting Principle.   This is more conveniently stated in words than as a formula, so formula-based studiers often overlook the importance of this idea, to their own detriment.   The more familiar counting ideas — more familiar because they have formulas associated with them!! — are permutations and combinations.  It’s important to recognize: all the ideas & formulas concerning permutations & combinations can be derived directly from the FCP, and in fact, often in the case of permutations, it’s much simpler to use the FCP directly than to use any permutation formula.

## Not so basic …

Those are the basic ideas.  They are relatively easy to state, and they are explained in the posts at those links.   The problem is: when you read an individual counting problem, how do you know which rule to use?

That’s precisely the hard part of counting, and what’s frustrating is that there’s not a simple flowchart/procedure we can recommend that will work for all counting problems.  The hardest thing about any counting problem is knowing how to begin, and essentially, this a right-brain pattern-matching procedure.  In the post How do to GMAT Math Faster, I discuss left-brain vs. right brain thinking in greater depth.   The gist is: any left-brain skill or procedure can be diagrammed and clearly explained in a logical step-by-step way, but any right-brain process is an inherently non-linear process into which one has to grow slowly, through experience.

Of course, in problem solutions, you can read about how they solved the problem, but this is tricky.    You see, in any counting problem, often the hardest part is how one initially perceives and dissects the situation, the way one breaks it into steps or stages.   Once that perceptual choice is made, especially if it’s a good choice, then applying any rules or formulas is easy.   If you just scan the solution for formulas & rules, you will say, yes, when we look it this way, of course we do that calculation.   That’s not what you need to study.  Above all, you need to study the very beginning: what was the very choice the solution made in solving the problem?  How did they begin the problem?  How did they frame the situation?  What were their perceptual choices?    Left-brain thinkers always want to know what to do, but in the right-brain realm, the most important choice is how to see, and once you are seeing things in the right way, it’s obvious what to do.  That’s especially what left-brain thinkers need to get from the solutions to counting problems.   Patterns emerge over time, and those initial perceptual choices become more and more natural.
Here are a few concrete recommendations:

1) Don’t just memorize the permutation & combinations formulas.  Memorization is not understanding.   Understand the FCP, and understand how that more fundamental idea underlies the formulas & procedures of permutations & combinations.   Those are the basic tools you need to know very well.

2) In counting problems, whether you get the problem right or wrong, always read the solutions.  Again, you are looking for how the solution began, the very first perceptual choice made before any calculations were done.  If you are more of a left-brain person, I actually would recommend keeping notes of the situation given in the question and the first few steps, the perceptual choices, made in the solution.   Forcing yourself to articulate the choices will help to build connections between the brain’s two hemispheres.

3) See the How do to GMAT Math Faster article for more tips on how to exercise and develop your right-brain.

## Summary

If you had trouble with the questions above, it may be that some of the information in this article or at one of the linked articles gives you a clue, but most likely you will have to study the solutions below.  Remember to be mindful of what you need to get from the solution to a counting problem, as discussed above.  Here’s another counting problem for more practice:

## Explanations for practice problems.

1) For those familiar with Kakuro, this problem presents a related challenge.  First, we need to know: how many combinations of four distinct single-digit numbers have a sum of 28?  Start with the three highest digits: 9 + 8 + 7 = 24, so {9, 8, 7, 4} is one such combination.  First, keep the 9 & 8, and bring the other two closer together: {9, 8, 6, 5}.  OK: that exhausts possibilities with 9 & 8 as the highest numbers.  Now, omit 8, and try 9 & 7: even with the largest available digits, 9 + 7 + 6 + 5 = 27, so this will not be big enough.  Similarly, without 9, even with the largest available digits, 8 + 7 + 6 + 5 = 26, also is not big enough.  If we omit either 9 or 8, the sum is not big enough, and with both 9 & 8, there are only two possibilities: {9, 8, 7, 4} or {9, 8, 6, 5}.   Thus, only two possible sets of four digits.

For each one of these, we could permutate the number in any order.  4! = 24, so for each of the two sets, there are 24 different four-digit numbers we could form.  That’s a total of 2*24 = 48 different four-digit numbers.

2) We will break the possible numbers into four different groups.  First, we will consider all the numbers with no repeat digits: how many different sets of four different number can I choose from the digits 1-6?  Remembering the tricks for calculating combinations, 6C4 = 6C2 = 15.  For each one of those, we can permutate the digits in any order: 4! = 24.   To multiply 15*24, use the doubling & halving trick.  Half of 24 is 12, and twice 15 is 30, so 15*24 = 30*12 = 360.  That’s group #1, all the four-digit numbers with four distinct digits.

Group #2 will be the group with two 6’s and the other two digits distinct.  Think about this.  There are 4C2 = 6 “slots” the two sixes could occupy:

6 6 _ _

6 _ 6 _

6 _ _ 6

_ 6 6 _

_ 6 _ 6

_ _ 6 6

Now, for each of those, use the FCP: we would pick five digits for the first empty slot (1, 2, 3, 4, 5), and that would leave us four choices for the second empty slot.  Thus, for each of those six, there are 20 four-digit numbers we could create. That’s 6*20 = 120 total in group #2.

Group #3 will be the group with two 5’s and the other two digits distinct.  Think about this.  For symmetry reasons, there must be exactly as many numbers in this group as there are in group #2.  There are 120 in this group.

Group #4 will have two 5’s and two 6’s constituting the four digits.   If we choose the two slots into which to drop the 5’s, the 6’s have to go into the other slots.  Therefore, there must be 4C2 = 6 of these.  Here they are: 5566, 5656, 5665, 6556, 6565, 6655.

The total is the sum across these four groups: 360 + 120 + 120 + 6 = 606

3) First, notice that in every square, we could construct four different triangle. In the above square, we can construct triangles ABC, ACD, ABD, and BCD, all congruent.

Well, if we take all the squares of this size in the diagram, We have 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 squares.  That’s 28 squares, times four triangles in each square: 28*4 = 112.

BUT, we can’t forget to count the leftover triangles on the diagonal slant: That’s 8 more triangles, for a total of 112 + 8 = 120

4) Well, first of all, ignoring the type of triangle formed, how many combinations total?  The easiest way to think about this is to use the Fundamental Counting Principle.   For the first dot, 15 choices, then 14 left for the second choice, then 13 left for the third choice: that’s 15*14*13.  But, that will count repeats: the same three dots could be chosen in any of their 3! = 6 orders, so we have to divide that number by 6.  (NOTICE the non-calculator math here).

(15*14*13)/6

Cancel the factor of 3 in 15 and 6

(5*14*13)/2

Cancel the factor of 2 in the 14 and 2

(5*7*13) = 5*91 = 455

That’s how many total triangles we could create.

Of these, how many are equilateral triangles?  Well, the only equilateral triangles would be three points equally spaced across the whole circle.  Suppose the points are numbers from 1 to 15.  From point 1 to point 6 is one-third of the circle — again, from point 6 to point 11, and from point 11 back to point 1.  That’s one equilateral triangle.  We could make an equilateral triangle using points

{1, 6, 11}

{2, 7, 12}

{3, 8, 13}

(4, 9, 14)

{5, 10, 15}

After that, we would start to repeat.  There are five possible equilateral triangles, so 455 – 5 = 450 of these triangles are not equilateral. ### 20 Responses to Difficult GMAT Counting Problems

1. Harveykl November 4, 2015 at 9:12 am #

All kinds of freaky formulas and rules to remember can make learning how to solve GMAT counting and combinatorics problems seem to be a major hassle. The truth is that by learning a little about how this all works, one can get this down no problem. So here is an outline that gets you to the more complex stuff while keeping it simple.

2. CK008 August 26, 2015 at 6:07 am #

Hello !
Thanks for the nice problems…. I am a left hander…. therefore right brained
How I start is:
1) What constraints do these problems impose
For Eg. question 1-
A) Sum of digits is 28 and no. is 4 distinct digit, therefore the average digit is 7 within max 9 and min 1, now with average 7 the play( possible digits) is only from 9 .i.e ( Max which is 7+2) to 5 i.e. (7-2)
B) _ _ _ _ : 9,8,_,_ => sum of other 2 digits is 28-(9+8)=11, options for sum 11 from remaining digits is 7&4, 6&5
C) Combination of 9,8,7,4 as well as Combination of 9,8,6,5
4!+4!
Numbers Game!

• CK008 August 26, 2015 at 6:17 am #

sorry C) step is permutation

3. Mike July 6, 2015 at 9:14 am #

Solution to problem 1:
checking numbers
1 x
2 x
3 x
4 ok (789)
5 ok (986)
6 ok (985)
7 ok (984)
8 ok (974)
9 ok (874)

So 6 Numbers are Ok –> 6! / (6!/4!2!) = 48

4. Binit March 11, 2015 at 12:40 pm #

Dear Mike,
Thanks a ton for suggesting this awesome post to me. This time I cud solve all of them (though Q.3 took a long time), I m happy to be able to see the right way to look at them.
Great explanations, as aways 🙂
Sincere appreciation,

Binit.

• Mike March 13, 2015 at 10:13 am #

Dear Binit,
You are more than welcome, my friend! Best of luck to you!
Mike 🙂

5. Hamad March 9, 2015 at 2:44 am #

Dear Mike,

I’ve confusions relating question # 2. While we are suppose to calculate distinct four-digit numbers that can be formed by digits (1,2,3,4,5,5,6,6), does it not mean that

We have 6 distinct digits i.e. 1,2.3,4,5,6

& so therefore, the distinct 4 digit numbers that can be formed should be

6*5*4*3 = 360.

• Mike March 9, 2015 at 11:54 am #

My friend, in all math problems, you have to be meticulous careful about the precise language. What must be “distinct” are the four-digit numbers. There is no claim that the digits should be distinct within the four-digit numbers. Indeed, since repeated digits are given in the set, that means we could have number such as: 5125, 3646, and 5665. Those are three distinct four-digit numbers, but they don’t have all distinct digits. You were applying a restriction given in only one aspect of the problem to other aspects as well.
Does all this make sense?
Mike 🙂

• Hamad March 13, 2015 at 9:28 am #

Dear Mike,

First of all, thankyou for such a quick response on my inquiry.

And yes, it does makes sense. It has indeed helped a lot.

I’ve heard from friends who appeared for GRE recently that there are tougher probability questions appearing on GRE now a days. I would like you to pleasr post some more difficult questions relating probability.

Thank you.

• Mike March 13, 2015 at 10:23 am #

I’m glad you found my response helpful. 🙂 As for your request, I want you to understand a subtle cultural difference. As someone who produces content, it takes me a long time to write question; good questions represent a great deal of hard work. When you simply ask, post more questions, its as if you are saying, do a great deal of hard work right now just for me. In America, this can be perceived as rude. Now, I totally realize, you were not intending any disrespect at all. I just want you to understand how this request lands on American ears. We all always have a lot to learn about other cultures. 🙂
Having said that, here are a few blogs with probability problems:
1) GMAT Probability Questions
2) GMAT Data Sufficiency Practice Questions on Probability
I would not recommend that 2nd set if you are not familiar with the GMAT question type of Data Sufficiency.
Mike 🙂

• Hamad March 13, 2015 at 12:30 pm #

Dear Mike,

I didn’t mean any disrespect at all. 🙂

I still appologise If I sound so.

And thank you once again for the efforts you put in for all of us to help us understand the concepts. 🙂 🙂

• Mike March 13, 2015 at 1:00 pm #

You are quite welcome, my friend! 🙂 No worries! I wish you the very best of good fortune!
Mike 🙂

6. Nurlan February 24, 2015 at 1:03 am #

I did all correct)) thanks gentleman. that is a good test.

• Mike February 24, 2015 at 3:01 pm #

Nurlan,
You are quite welcome! Congratulations on good work!
Mike 🙂

7. alysha November 14, 2013 at 5:04 am #

Hello Mike, I am impressed to view the answers. It’s really amazing and I never did this type of math before. But from now I will follow you as well as magoosh to increase my math knowledge. Thanks.

• Mike November 14, 2013 at 9:32 am #

Dear Alysha,
Thank you for your compliments. Best of luck to you.
Mike 🙂

8. KConfused November 7, 2013 at 1:26 am #

Question 3: Does congruence mean equal sides? Or equal proportions of sides?
If it is proportions, shouldn’t we consider bigger squares too? (Of course with similar side to side ratios)

Thanks
KC

• Mike November 7, 2013 at 9:51 am #

Dear K Confused,
The word “congruent” means “same shape, same size”. The word for “same shape, different sizes”, with equal proportions, is “similar”. See:
https://magoosh.com/gmat/2013/gmat-math-similar-shapes/
Mike 🙂

9. prateek sharma November 5, 2013 at 4:50 am #

Last question was very tough !!

I hope I do not encounter such questions on GMAT . 🙂

• Mike November 5, 2013 at 9:49 am #

Prateek,
These are very hard questions. As I said above, you would have to get a great deal of other math perfectly correct to land in this zone of the CAT. At least you’ve see it once here! Good luck, my friend.
Mike 🙂

Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.

We highly encourage students to help each other out and respond to other students' comments if you can!

If you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!