Each of the five countries sends a delegation of three representatives

Question

Each of the five countries sends a delegation of three representatives to a conference. How many three-person committees could be chosen that do NOT include more than one representative from a single country?

A. 156
B. 270
C. 455
D. 910
E. 2730

Show Answer

Official Answer

B

A. 156
B. 270
C. 455
D. 910
E. 2730

Bunuel · Accepted Answer

I think you might be missing an important point. Here’s the solution again:

Each of 5 countries sends 3 people to a conference. How many 3-person committees could be chosen that do not include more than 1 representative from a single country?

A. 156
B. 270
C. 455
D. 910
E. 2730

We have 3 people from each of the 5 countries:

{1, 2, 3}
{4, 5, 6}
{7, 8, 9}
{10, 11, 12}
{13, 14, 15}

We want to form 3-person committees that do not include more than 1 representative from a single country.

First, we need to choose 3 countries that will each provide 1 representative for the committee. This can be done in 5C3 ways, which equals 10 (the number of ways to select 3 countries from 5).

Now, consider one set of 3 countries chosen, for example:

{1, 2, 3}
{7, 8, 9}
{13, 14, 15}

We want to choose 1 person from each group. Since there are 3 options in each group, the total number of ways to choose one person from each country is 3 * 3 * 3.

So, the overall number of ways to form the committee is 5C3 * 3 * 3 * 3 = 270.

Answer: B.

Hope it's clear.

woohoo921 · Answer

GMATNinja did 5!/3!2!=10 ways to choose 3 countries and then proceeded to calculate 10*3*3*3=270. I am a bit confused on the steps here. Any clarification/another way to look at the problem would be greatly appreciated

MartyTargetTestPrep avigutman egmat

youcouldsailaway · Answer

There are in total 5 countries that can each send one person as a representative. So in terms of selecting the number of people who can join the committee, we can select 3 different people from 5 different countries. Thus (given I can only select 3 people) my set here is 5C3, which is 10.

Now, the total number of people being sent from each country is 3. So I can choose one person from the set of 3 sent by each country. Hence my set here is 3^3 = 27.

Conversely, 3C1 X 3C1 X 3C1 = 27.

Combining the two above to get the total possible set = 10*(3^3) = 270.

egmat · Answer

Hey  woohoo921 , 
Thanks for posting your query.

Let me help you understand the approach that GMATNinja used. That is my chosen approach, too!

UNDERSTAND THE QUESTION    
Let’s start writing everything that is given to us in the question. We will keep making inferences as we go ( Always advised!  - Do not just keep reading the question without   translating   and   inferring   as you go. This will save you from getting overwhelmed.)

Okay, let’s start. We have: 
  5 countries (say, A, B, C, D, E); Each country sends 3 people to a conference. ---- (1)  
 A 3-person committee is to be chosen 
  These 3 people do  not  include more than one person from any one country.  
 To elaborate, we  CANNOT  have the following: 
  All 3 members from a single country, or  
 2 members from a single country and the third from another.   
 So, we can   infer   that each member of the 3-member committee comes from a different country. ----- (2)

SOLUTION    
Now that we understand the question completely, we will also be able to understand the solution properly. Let’s go!

From (2), we know that each member of the 3-member committee comes from a different country. 
  So, we first need to select the 3 countries from which these 3 committee members come.  
 Out of a total of 5 countries, in how many ways can we select 3? 
  We can do this in  ^5C_3  =  \frac{5!}{(5−2)!}  = 10 ways. ------------(3)     
What’s next now? Let’s understand through an example. 
  Out of the 10 possibilities obtained in (3), let’s consider one. Suppose A, C, and D are the 3 countries from which the 3 committee members come.  
 Next, we need to find which person from A, C, and D each became the committee member. 
  We know from (1) that like every other country, each of A, C, and D sends a total of 3 members for the conference.  
 So, for country A, we need to select 1 out of the 3 people it sends. This can be done in  ^3C_1  = 3 ways. -------(4)  
 Similarly, for country B, we select 1 out of 3 people in  ^3C_1  = 3 ways. ----(5)  
 And for country C, we again select 1 out of 3 people in  ^3C_1  = 3 ways. ----(6)   
 Finally, combining (3) through (6), we conclude that the required answer = 10 × 3 × 3 × 3 = 270.

TAKEAWAYS    
  Read the question piece by piece and keep translating English to Math as you go. 
 Keep making inferences as you gather more and more information while reading questions or solving them. 
 Take simpler examples, wherever possible, to understand what exactly you need to do in a question.

Hope this helps!

Best, 
Shweta 
Quant Product Creator, e-GMAT

SaquibHGMATWhiz · Answer

Solution:

Number of ways of choosing 3 countries involved in the selection  = 5C3=10  
 Now, from every country, we have 3 people to choose from 
 So, the number of ways to choose that  =3	imes 3	imes 3=27  
 Thus, total number of ways  =10	imes 27=270

Hence the right answer is    Option B

mysterymanrog · Answer

This is how I approached the question:

If each country sends 3 people, and there are 5 countries, then the total number of people to choose from is 15.

To pick the first member, we can select anyone: 15choose1=15
To pick the second member, we must remove the person already selected, and the 2 other people from the same country as the selected person: 12choose1=12
To pick the third member, we repeat the same as the process to pick the second member: 9choose1=9

That means we have a total of: 15*12*9 ways to choose.

However, we have to remove 3! arrangements, because the groups in this situations are not unique - so it is \(\frac{15*12*9}{3!}=270.\)

Here is a similar GMATPrep question to this, in which you can use the method I used. https://gmatclub.com/forum/a-committee- ... 94068.html

SignUp · Answer

Okay this really isnt a solution but a way to guess the answer when you can't think of a way out.

So the maximum possible ways to select a committee of 3 out of 15 people (without any conditions whatsoever ) is - 15C3 = 455

So 455 is the maximum number of ways to select so C,D and E are eliminated

Out of A and B considering we have a 5 (number of countries) answer will most probably be something divisible by 5 .

Hence B

RushilVasudev · Answer

This is a logical question! Not include cases = Total - Included cases

Total = 15C3 = 455, Inclduded case = No need to find

We know the answer has to be less than 455. Out of B (270) and A (156), A can't be the answer as it is too low. Hence B

GKLemon · Answer

This question can be solved by considering this equation:

15C3 - ((5C1 x 3C2) x (4C1 x 3C1) + 5C1) = 270

or it's simplified counting form

15C3 - (5 x 3 x 4 x 3 + 5) = 270

To understand why we need to break down what this question is giving and asking us first:

Each of the five countries sends a delegation of three representatives to a conference.

That means there are 5 countries, picking to send a delegation of 3 representatives to a conference; We can infer from this statement that each of the 5 countries has 3 representatives to pick from.

This brings the total number of representatives to choose from to 15.

How many three-person committees could be chosen that do NOT include more than one representative from a single country?

the question is delivered here: How many  three-person  ........ chosen ..... that do  NOT include more than one rep  from a  single country

translating the mysteriously confusing question here into normal English:

Picking from a pool of 15 reps from 5 different countries, whereas each country has 3 reps:

How many combinations of 3 reps can we pick from those 15 reps whereas (the qualifier) the delegation sent does not include more than 1 rep from each country?

Translating the above to mathematical expression:

Total Combination of Delegates - Combination that includes 2 or 3 rep from the same country = the answer

Total combination of Delegates  = 15C3 (Out of 15 delegates, choose 3) = 455

Combination that includes 2 or 3 reps from the same country  =

2 reps: There is only  5C1  way = 5 choices (since there are only 3 delegation spots if 1 country has 2 reps, then there are only 5 possible ways to pick 2 reps from the same country, as 2 reps from the same country can only be slotted into 3 spots in 1 way per country)
There are  3C2  ways to choose these 2 reps, 3C2 = 3
so,  the ways to pick 2 reps from the same country = 5C1 x 3C2

And following the above, we have 4 ways to choose the last delegate from any of the 4 remaining countries that do not have 2 reps this is 4C1
There are 3c1 ways to choose this last rep, 3C1 = 1
so,  the ways to pick 1 rep from the remaining 4 countries that did not send 2 delegates = 4C1 x 3C2

And there are 5 ways to choose a delegation where all three reps are from the same country. 5C1

Putting it all together:

Combination that includes 2 or 3 reps form the same country =  (5 x 3) x (4 x 3) + 5 = 185  =  Combination that includes 2 or 3 reps form the same country

To break down the above

(5 x 3) = ways to choose 2 delegates from the same country out of 5 countries:
(4 x 3) = ways to choose 1 delegate from the leftover countries that didn't send 2 delegates
+5 = ways of sending 3 delegates from the same country.

Now we take the total - the exception

= 455 - 185 = 270

Answer is B

satkeerti · Answer

Why is it wrong to say:

15C1 x 12C1 x 9C1 ?

we have 15 options to choose the first person, 12 choices for the second and 9 choices for the third.

Bunuel · Answer

The approach of using combinations to select each representative one at a time (15C1 x 12C1 x 9C1) would lead to overcounting of the possible combinations, as it will give ordered groups.

For instance, consider a situation where we have representatives A1, A2, A3 from Country A and B1, B2, B3 from Country B, and C1, C2, C3 from Country C.

Using the 15C1 x 12C1 x 9C1 approach, you'd get A1, B1, C1 group as well as B1, C1, A1, group, even though those are the same groups of three.

To get the correct answer, you should divide 15C1 x 12C1 x 9C1 by 3! to get the number of unordered groups: 15C1 x 12C1 x 9C1/3! = 270.

The correct solution 5C3 x 3 x 3 x 3 does not overcount because it first chooses the 3 countries (5C3) that will provide a representative, and then selects one representative from each of those 3 countries. This approach assumes that the committee of A1, B1, C1 is the same as B1, C1, A1, which is correct as per the problem's requirements.

Hope it helps.

SethLowell · Answer

Thats really nice, but i still dont get it why all solutions look at one committee if the question asks for the multiple of committees.

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Each of the five countries sends a delegation of three representatives

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