# Lesson: GMAT Counting Strategies - Part II

## Comment on GMAT Counting Strategies - Part II

### Hi Brent,

Hi Brent,

How we should go about with problems that involves identical objects and the outcome of one stages does not differ from the outcome of other stage i.e. it is a MISSISSIPI Rule + Combination problem?

### Hmmm, I'm trying to think of

Hmmm, I'm trying to think of a question that meets these conditions. The problem is that the MISSISSIPPI rule implies that the order of the letters matters, whereas we typically use combinations when the order doesn't matter.

Do you have an example question in mind?

### http://gmatclub.com/forum/in

http://gmatclub.com/forum/in-how-many-different-ways-can-3-identical-green-shirts-and-105384.html

This is question that involves identical objects and where order does not matter. Generally the solutions employ MISSISSIPI Rule. Since the order does not matter in the question, there is one solution that solves it with Combination technique (however it is appears quite tricky). Hence, I am unsure how to go about in such cases. And how these both techniques are related to each other.

### That question (http:/

That question (http://gmatclub.com/forum/in-how-many-different-ways-can-3-identical-gre...) is unique in that we can look at it in two different ways.

We can think of it as ordering the letters RRRGGG such that child #1 gets the color indicated by the 1st letter, child #2 gets the color indicated by the 2nd letter, etc. In this approach, we can apply the MISSISSIPPI rule.

Alternatively, we can think of it as simply CHOOSING which 3 children receive a red shirt (and the remaining 3 children receive the green shirt). In this approach, you can use combinations.

This really shouldn't pose a problem for you. Instead, it's a bit of a gift, since we can use EITHER method to solve it. That's a good thing :-)

### Hi Brent, first of all,

Hi Brent, first of all, thanks a lot again for these brilliant videos, I find them very helpful!

But I do wonder where I'm wrong with this line of reasoning:
- After Stage One, we have 3 married couples (6 people) to choose from.
- I then proceeded to calculate the rest by using the FCP:

Since we want to choose 3 people, and the 3 selections are equally restrictive (I think? the only restriction being "not married to another person on the committee") we can choose them in the following order: ___1st___ , ___2nd___ , ___3rd___

- There are 6 ways to choose the first person, and since his/her spouse cannot be chosen after making the first selection, there will be 4 more people to choose from for the second position, and applying the same logic, there will be 2 ways left to select the third person, so using the FCP I calculated 6x4x2..

Which, of course, yielded a wrong answer, but I can't seem to see the flaw in my line of reasoning. Maybe you can help shed some light on this issue?

Many thanks,
Kai

### The only problem with your

The only problem with your approach is that it treats the outcome of stage 1 as different from the outcome of stage 2.

Let's represent the 6 people as A, a, B, b, C, c (where A and a are a couple, as are B and b etc)

In your approach, selecting A then b then C is considered different from selecting b then C then A, when they both yield the same committee.

In fact, your approach takes each arrangement and counts each of them 6 times. For example, in your approach, the following committees are considered different: bCA, bAC, AbC, ACb, CAb, CbA

NOTE: there are 3! ways to arrange A, b and C.

So, to account for all of this duplication, we must take your result of 6x4x2 and divide it by 6 (aka 3!). if you do this, then you will get the correct answer.

### Dear Brent,

Dear Brent,

I solved the "In a meeting of 3 representatives from each of 6 different companies, each person shook hands with every person not from his or her own company. If the representatives did not shake hands with people from their own company, how many handshakes took place?" Question using the subtracting the unwanted results rule.

Total number of Handshakes possible = 18C2 = 153
Total number of same company hanshakes = 6 companies x 3C2
= 6 x 3 = 18
Therefore no of other company handshakes = 153 - 18 = 135

That's a perfectly valid approach, aashaybaindur - well done!

### Brent,

Brent,

Could you explain me the exercises #169 & #207 from OG17 using the restriction rule?

Thanks
Pedro

### Hi Pedro,

Hi Pedro,

Those questions don't lend themselves to using the restriction rule (especially question #207).

Cheers,
Brent

### Hi Brent,

Hi Brent,
This question is in the practice questions given below.But i am not able to see any expert reply on this.
Can you please give me solution of this question.

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Thanks

### Hi Brent, I have a doubt? In

Hi Brent, I have a doubt? In this question: https://gmatclub.com/forum/jack-is-making-a-list-of-his-5-favorite-cities-he-will-choose-3-citie-239856.html you solved the problem this way: 5C3 * 3C2 * 5!, I understood the first two stages but not the last one. If I multiply per 5! I would not take the risk of set a city which was already set in the first two stages?

In stages 1 and 2, I am selecting my 5 cities (3 US cities and 2 European cities). So, let's say the 5 cities are: Seattle, Chicago, Miami, Paris and Madrid.

However, at this point, all we have done is select the 5 cities; we have not yet RANKED them from 1st to 5th (as the question directs us to do).

So, for example, some possible rankings are:
- 1st-Seattle, 2nd-Paris, 3rd-Madrid, 4th-Chicago, 5th-Miami
- 1st-Chicago, 2nd-Miami, 3rd-Madrid, 4th-Seattle, 5th-Paris
- 1st-Miami, 2nd-Chicago, 3rd-Paris, 4th-Seattle, 5th-Madrid
etc

So, we must still count the various ways we can rank the cities (5! ways). We handle this task in stage 3.

Does that help?

Cheers,
Brent

### Hi Bret,

Hi Bret,

Can you please explain the solution to the following problem you posted on GMATCLUB? I dont understand the solutions that were provided by other experts.

Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?

A) 36
B) 45
C) 55
D) 66
E) 78

Can we see a problem of this magnitude on the GMAT? This seemed really hard. I would've guessed and moved on and not wasted my time on it. But I am def curious to know if there is an easier solution to this.
Thank you,

Ari

I was quite proud of that question UNTIL someone pointed out a serious flaw: If each team plays every other team, then it's impossible for 2 or more teams to win EVERY game, since those teams would have played each other, which means one of the teams had to lose.

That said, if we ignore the flaw, it's a very time-consuming question, AND it's a 750+ level question. So, guessing and moving on would be a good idea.

Here's the solution I posted on a different forum BEFORE someone pointed out the flaw in my question.

### Hi Brent,

Hi Brent,

Could you please show a humanly way of solving this Qs? I dont understand the explanation on Gmatclub at all.

Its an OG question to top it all so I am really worried that I dont get the concept.

A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

https://gmatclub.com/forum/a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html

I've created two different solutions to this question.

Cheers,
Brent

### Hi Brent :) I don't

Hi Brent :) I don't understand exercise 201 in the gmat official guide 2019. Can you give me your solution?

### My solution: https://gmatclub

Aside: Here are links to solutions to ALL official guide questions: https://gmatclub.com/forum/gmac-official-guides-the-master-directory-lin...

Cheers,
Brent

### Hi Brent ,

Hi Brent ,

I needed some help with 2 questions :

1. https://gmatclub.com/forum/m06-183708.html : Why did we divide it by 4! here ? Shouldnt it be multiplied instead ?

2. https://gmatclub.com/forum/m26-184458.html : Like the 3 sixes can assume any positions within the 5 numbers , Cant the 2 numbers interchange the position between themselves , shouldn’t we multiply it by 2 ? like 66465 and 66564 ?

Thankyou !

### 1. https://gmatclub.com/forum

1. https://gmatclub.com/forum/m06-183708.html
I'm not a big fan of this question, because it's unclear whether the 4 teams are considered the same or different.
For example, is Team 1 = A&B, Team 2 = C&D, Team 3 = E&F, Team 4 = G&H different from Team 2 = A&B, Team 1 = C&D, Team 3 = E&F, Team 4 = G&H?
Since the official answer is B, we know that the INTENT of the question is to consider the two above configurations as equal.

In Bunuel's solution, he first treated the configurations as different, which means he counted many configurations more than once.
In fact, we can take the 4 different pairs and arrange them in 4! different ways (24 ways).
This means the first part of Bunuel's solution counted each configuration 24 times.
To account for this, he divided that value by 4! (aka 24)

2. https://gmatclub.com/forum/m26-184458.html
I'd have to see your full solution to determine whether you need to multiply your answer by 2.
In the meantime, here's my full solution: https://gmatclub.com/forum/m26-184458-20.html#p2301977

Cheers,
Brent

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-a-committee-of-3-people-is-to-be-selected-from-among-88772-20.html#p1729049

In this question. I am looking to calc it by Total-no. of ways to break it
Total ways is 10C3 = 120 - No. of ways committee includes 2 members who are couple

I then looked to use FCP _ _ _ = 10*1*8 = 80

The problem with your solution is that, for the number of ways to BREAK the restriction, you're counting each outcome TWICE.

You are saying there's 10 ways to select any one person.
Then, in order to break the restriction, you next choose that person's partner, and this can be done in 1 way.

So, for example, let's say A and B are a couple,
In your approach, we might choose A first and then choose B second.
Or, we might choose B first and then choose A second.
Your solution treats these two outcomes as DIFFERENT, when they are clearly the same.

Rather than choose 2 people from the same couple, just choose 1 COUPLE.

We can choose 1 COUPLE in 5 ways.
We can then choose 1 of the 8 remaining people in 8 ways.
So, the total number of ways to break the restriction = (5)(8) = 40

Correct answer = 120 - 40 = 80

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,
https://gmatclub.com/forum/what-is-the-number-of-the-shortest-routes-from-x-to-y-through-z-233292.html

Instead of stopping at Z to calculate the first 4 steps, I calculated from the start till the end, wherein I got 3Rs 3Us (both ways by starting from the right side). So I calculated 6!/3!3! from which I got 20 instead of 12. Can you please help me understand where am I wrong here?

Thank you!

That's a good idea. The only problem is that not all arrangements of 3 R's and 3 U's will pass through point Z.
For example, RRRUUU gets to point Y, BUT the path does not pass through point Z.

Likewise, RRURUU does not pass through point Z.
And URUURR does not pass through point Z.
etc

Cheers,
Brent

### Hi Brent, Q DS06659 of OG

Hi Brent, Q DS06659 of OG 2019 -
"S is a set of points in the plane. How many distinct triangles can be drawn that have 3 of the points in S as vertices
(1) there number of distinct points in S is 5.
(2) no three of the points in S are collinear."

Here the official answer is C, but shouldn't it be E because one of the factors of making a triangle is that sum of any 2 sides is greater than the 3rd side. So if there are any 3 points with distance 1, 2 and 3, in that case a triangle cannot be made. This way possible answers could be 9 (10-1).

### That's a good idea, Kashaf,

That's a good idea, Kashaf, but there aren't 3 such points so that the three lengths are 1, 2 and 3 (if try to find three points that meet that condition, you'll find that it's impossible to do so).

As long as we have three non-collinear points, we can always connect them to create a triangle.

Cheers, Brent

### hello sir!

hello sir!
greeting of the day!
needed your help with this question and is this type of question expected on the test day?

https://gmatclub.com/forum/each-in-the-mileage-table-above-represents-an-entry-indica-95162.html

thank you
tanya

### Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/each-in-the-mileage-table-above-represents-an...

This is an official (retired) GMAT question, so it's well within the scope of the test.

Cheers,
Brent

### hello sir,

hello sir,
greetings of the day!

Can you provide a better solution for this question-
https://gmatclub.com/forum/a-researcher-plans-to-identify-each-participant-in-a-certain-medical-134584.html

thank you,
Tanya

### I have two different

I have two different solutions to this question: