Lesson: GMAT Counting Strategies - Part II

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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?
gmat-admin's picture

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-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.
gmat-admin's picture

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, thanks a lot again for these brilliant videos, I find them very helpful!

So, after reading your post, I understand your solution/ explanation here: http://www.beatthegmat.com/if-a-committee-of-3-people-is-to-be-selected-t292338.html

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
gmat-admin's picture

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,

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
gmat-admin's picture

Question link: https://gmatclub.com/forum/in-a-meeting-of-3-representatives-from-each-o...

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

Brent,

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

Thanks
Pedro
gmat-admin's picture

Hi Pedro,

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

Cheers,
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
gmat-admin's picture

Thanks

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?
gmat-admin's picture

Question link: https://gmatclub.com/forum/jack-is-making-a-list-of-his-5-favorite-citie...

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

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