sam2304 wrote:A committee of three people is to be chosen from the president and vice president of four di¤erent companies. What is the number of di¤erent 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

Let's break the task into stages and then apply the Fundamental Counting Principle (FCP)

Stage 1: Select the 3 companies from which the committee members will be drawn.
There are 4 companies, so this stage can be accompanied in 4C3 ways (4 ways)

Stage 2: From one of the 3 chosen companies, select either the President or Vice President.
This stage can be accompanied in 2 ways.

Stage 3: From another of the 3 chosen companies, select either the President or Vice President.
This stage can be accompanied in 2 ways.

Stage 4: From the last of the chosen companies, select either the President or Vice President.
This stage can be accompanied in 2 ways.

So, the total number of ways = [spoiler]4x2x2x2 = 32 = D[/spoiler]

To learn more about the Fundamental Counting Principle, check out this video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

sam2304 wrote: I followed this approach.
4 companies and 8 people totally to choose from, 4 president and 4 vice president for each company. So total way to choose 3 people is 8 ways for 1st one, neglect the other one from the same company so 6 ways for second and 4 ways for third one = 8*6*4. This is not among the answer choices. Where am I going wrong ?

Your approach is almost perfect, except for one problem.
To show the problem, I'll follow your steps:

8 ways for 1st one.
Let's choose Al from company A.

Neglect the other one from the same company so 6 ways for second
Let's choose Bob from company B.

4 ways for third one
Let's choose Carol from company C.

This would count as one selection.

Let's try that again.

8 ways for 1st one.
Let's choose Bob from company B.

Neglect the other one from the same company so 6 ways for second
Let's choose Al from company A.

4 ways for third one
Let's choose Carol from company C.

With your approach, these two committee selections are treated as different, when they are actually the same.

To account for all of this duplication, we need to divide your solution (8*6*4) by the number of repeated solutions you have.

Since every 3-person can be arranged in 3! ways, we need to divide (8*6*4) by 3!. When we do so, we get [spoiler]32 (D)[/spoiler]

Cheers,
Brent