A committee of 3 people is to be chosen from the president

Question

A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40

A) 16
B) 24
C) 28
D) 32
E) 40

Bunuel · Accepted Answer

Each company can send only one "representative" to the committee. Let's see in how many ways we can choose 3 companies (as there should be 3 members) to send only one "representatives" to the committee: 4C3=4. But these 3 chosen companies can send two persons (either president or vice president ): 2*2*2=2^3=8. Total # of ways: 4C3*2^3=32. Answer: D. Or: 8*6*4/3!=32, we are dividing by 3! because the order in the committee is not important. Answer: D. Similar problems with different approaches: combination-permutation-problem-couples-98533.html ps-combinations-94068.html committee-of-88772.html Hope it helps.

pinchharmonic · Answer

man bunuel always has such a quick and elegant way. Can you comment if this approach is correct?

i just did the total number of possibilities first:

ab cd ef gh are the 4 companies for example

8c3. this will include the problem counts of a pres/vp of the same company

how many ways did we over count = How many ways can a president and VP of the same company be part of this group?

choose the group, 4c1, then take both the president and vp, 2c2, then a remaining member from one of the other 3, 6c1

4c1 * 2c2 * 6c1
4 * 1 * 6 = 24

8c3 - 24 = 56 - 24 = 32

Bunuel · Answer

Yes, that's also a correct approach.

mygmatsuccess · Answer

Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.

Bunuel · Answer

Yes. This is the same exact approach given in my first reply above.

KanishkM · Answer

Now this is quite amusing, i would consider the P & VP to be a couple

So there will be 4 couples, out which we need a committee which wont have one Total number of ways, without any restriction = 5C_3 = 56 ways Now if we keep a restriction we can calculate the ways in which one couple is present = 4C_1 * 6C_1 = 24 ways Now the difference will give us 32 ways

luisdicampo · Answer

Deconstructing the Question  
There are 4 companies, each with a president and a vice president.
Total people = 8.
We must choose a committee of 3 people, with the restriction that no two people can be from the same company.

Step-by-step  
Since no two can be from the same company, we must select 3 different companies.

Number of ways to choose 3 companies from 4:
 C(4,3) = 4

From each chosen company, choose 1 of 2 people:
 2^3 = 8

Total number of committees:
 4 × 8 = 32

Answer: 32 (D)

repellatquas · Answer

Bunuel  why is this approach incorrect -
Choosing one company out of 4 -> 4C1*2C1 (since any one out of the two can be selected)
Similarly, choosing one company out of the remaining 3 -> 3C1 *2C1 
Finally, choosing one company out of the remaining 2 -> 2C1*2C1

Total = 4*2 + 3*2 + 2*2 = 8+6+4 = 18

Bunuel · Answer

Two issues.

The steps are sequential selections, so they must be multiplied, not added. Adding treats them as separate alternatives, which is incorrect. 
 Even after multiplying, the same committee is counted multiple times because the order of choosing the three people is irrelevant. So the result must be divided by 3!.

repellatquas · Answer

Understood! Can you also please state where and how I have chosen order here? Is it not simply taking 1 out of 4 companies?

Bunuel · Answer

The order appears because the method selects the companies one step at a time: first from 4, then from the remaining 3, then from the remaining 2.

This creates ordered selections of the three companies. The same set of three companies can be formed in multiple orders, which is why that approach overcounts unless you divide by 3!.

GMATinsight · Answer

Choose 3 companies for choosing 3 people from = 4C3 = 4 ways

Now, from each company one person may be chosen in 2 way i.e. outcomes = 2*2*2

So total Outcomes = 4C3*2*2*2 = 32

Answer: Option D

Related Video:  https://www.youtube.com/watch?v=I2XaJ2qEV1E

A committee of 3 people is to be chosen from the president

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