Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.

- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos

- Study Guide
- Philosophy
- Office Hours
- Extras
- Prices

## Comment on

Committees without R and S## Why are you making 4 person

## Good catch! Funny, you're the

Good catch! Funny, you're the first person to point that out. We'll fix that video shortly.

## How many possible questions

## They aren't that common. On

They aren't that common. On test day, you might see 1 or 2 counting questions.

## What should the real answer

## With a 3 person committee,

With a 3 person committee, the answer is 30 (35 possible 3-person committees - 5 committees that break the rule)

## Why can't we do 5C3 in the

## Good question, Ketan.

Good question, Ketan.

I think I understand your rationale for wanting to use 5C3.

You're saying that, if Rani and Sergio cannot be on the same committee, just remove them and select 3 people from Takumi, Uma, Vivek, Walter and Xavier.

We can do this in 5C3 ways.

The problem with that approach is that it doesn't allow for certain outcomes.

The restriction that Rani and Sergio cannot be on the same committee still allows for ONE of them to be on the committee.

For example, an acceptable committee is: Rani, Walter and Xavier

However, your 5C3 approach doesn't allow for that configuration.

Does that help?

Cheers,

Brent

## Thanks! I now get it :)

## I will learn the Restrictions

I thought I had 3 possible scenarios.

1) Neither Rani or Sergio are on the committee

2) Rani is on the committee

3) Sergio is on the committee

For

1) I said since R and S are out, we have a 5Choose3 = 10

2) R is picked, so S is out. So, first slot only has 1 choice. Second choice has only 5 choices left because R and S are out. Third choice, only 4 remain. So, putting it all together (1x5x4)/3! = 20/6

3) S is picked, so R is out. So, first slot only has 1 choice. Second choice only 5 choices left. So, (1x5x4)/3! = 20/6

If I sum the outcomes 1-3, I get 22 and 6/20, which is obviously wrong.

Can you tell me why the approach I used is wrong?

## For case #2, we can first

For case #2, we can first assign Rani to the committee.

Now we must choose 2 more people for the committee.

There are 5 people to choose from.

So, we can add the other two people in 5C2 ways (= 10 ways)

NOTE: You would have also got 10 in your solution had you taken (1x5x4) and divided by 2! (since choosing R was already a given)

The same rationale applies to case #3

Cheers,

Brent

## Regarding the note: I thought

So, if I had a lottery of 5 numbers picked from 25, and two numbers were predetermined, then the number of combinations would be equal to 23C3?

1x1x23x22x21/(r - predetermined choices)!

1x1x23x22x21/(5 - 2)!

1x1x23x22x21/(3)!

## It's easier to show the

It's easier to show the problem if we use smaller numbers.

Let's say there are 3 people (R, S and T) and we want to create a 2-person committee such that R and S are not on the same committee.

We'll examine 2 cases:

1) R is on committee, and S is not on committee

2) S is on committee, and R is not on committee

Case 1)

Using your approach, R is picked, so S is out.

So, the first slot only has 1 choice.

The second slot has only 1 choice left (since S is out)

So, putting it all together (1x1)/2! = 1/2

As you can see, it makes no sense to divide by two in this case.

-------------------------------------

Having said that, here's a way you can still use your strategy:

I'll refer you to your approach regarding case 2 (where R is picked, but S is not)

Rather than say "first slot only has 1 choice," you can choose which of the 3 slots you want to place R.

From here, you can fill the first available slot with 5 choices (since R and S are out).

The last available slot can be filled with 4 choices

We get: (3x5x4)/3! = 60/6 = 10

Cheers,

Brent

## or maybe the division should

## That would work in this case.

That would work in this case.

Alternatively, the applying the Restrictions Rule would be the best option.

## Thanks a lot. I appreciate

## For the breaking restrictions

## We can't use the FCP here,

We can't use the FCP here, because the outcomes for stages 1 and 2 (in your approach) are the same.

In your solution, you're suggesting that choosing Sergio and then Rani to be on the committee is different from choosing Rani and then Sergio to be on the committee. However, these two outcomes are the same.

Does that help?

## For this question I

## I'm not sure how you're

I'm not sure how you're getting (6*5*4)/(3!) for the number of ways to break the condition.

Once you put R and S on the committee (to break the condition), you now have 2 members of the 3-person committee.

So, at this point, we need only choose 1 more person from the remaining 5 people.

Also keep in mind that, the glue method is used for arrangements in which the order matters (e.g., people seated in a row), whereas this question involves committees in which the order of the members does not matter.

## https://gre.myprepclub.com/forum

To elaborate on my last question in this specific problem we used the glue method (for the break part) can you explain why for this one we did and for the previous comment I submitted why we did not?

## For this question, the order

For this question, the order in which the letters appear matters, whereas the the order of the committee members does not matter in the question involving Rani and Sergio.