Have questions about your preparation or an upcoming test? Need help modifying the Study Plan to meet your unique needs? No problem. Just book a Skype meeting with Brent to discuss these and any other questions you may have.

- Video Course
- Video Course Overview
- General GMAT Strategies - 7 videos (free)
- Data Sufficiency - 16 videos (free)
- Arithmetic - 38 videos
- Powers and Roots - 36 videos
- Algebra and Equation Solving - 73 videos
- Word Problems - 48 videos
- Geometry - 42 videos
- Integer Properties - 38 videos
- Statistics - 20 videos
- Counting - 27 videos
- Probability - 23 videos
- Analytical Writing Assessment - 5 videos (free)
- Reading Comprehension - 10 videos (free)
- Critical Reasoning - 38 videos
- Sentence Correction - 70 videos
- Integrated Reasoning - 17 videos

- Study Guide
- Your Instructor
- Office Hours
- Extras
- Prices

## Comment on

Committees without R&S## How can we solve this without

## We would have to examine

We would have to examine three cases:

1) Rani is on the committee but Sergio is not

2) Sergio is on the committee but Rani is not

3) Neither Rani nor Sergio is on the committee

1) First place Rani on the committee, and then select the two other committee members from Takumi, Uma, Vivek, walter and Xavier. We can select the 2 members from the 5 remaining people in 5C2 ways (= 10 ways)

2) First place Sergio on the committee, and then select the two other committee members from Takumi, Uma, Vivek, walter and Xavier. We can select the 2 members from the 5 remaining people in 5C2 ways (= 10 ways)

3) Select all three committee members from Takumi, Uma, Vivek, walter and Xavier. We can select the 3 members from the 5 eligible people in 5C3 ways (= 10 ways)

So the TOTAL number of committees = 10 + 10 + 10 = 30

## I tried a different approach

thanks much!

## It's hard to tell whether

It's hard to tell whether that's a valid approach. Can you tell me what the 3C1 is referring to?

## On listing the committees (at

## Good catch! It should be

Good catch! It should be groups of 3 (as per the question) We'll fix that shortly.

## Hi Brent,

I was trying to think out of the box but dont know where I went wrong !

Case 1: lets remove Rani from the list...we are left with 6 ppl (AND RANI AND SERGIO WILL NOT BE TOGETHER)

Now, we can have 6C3 ways of selecting a 3-people committee (with RANIand SERGIO not together)

Case 2 : Remove Sergio from the list...

Now, there are 6C3 ways to make a committee.

Add them up we get 20+20 ways = 40 ways.

Can you please suggest where I am wrong in here.

Thanks in advance.

## That's a good start, but your

That's a good start, but your approach ends up counting some outcomes more than once.

For example, in your case 1, we remove Rani from the list. After doing so, there are 6 people remaining, from which we will select 3 people. This can be done 20 ways. Notice the 10 of these 20 outcomes include NEITHER Rani NOR Sergio. For example, some outcomes that include NEITHER Rani NOR Sergio are {T, U, V}, {T, V, X} and {U, V, W}

I know there are 10 outcomes that include NEITHER Rani NOR Sergio, because if we want to create such groups (containing neither R nor S), we can select any 3 people from T, U, V, W and X. We can select 3 people from these 5 people in 5C3 ways (= 10 ways)

Likewise, in your second case, we remove Sergio. Once again there are 6 people remaining, from which we will select 3 people. This can be done 20 ways. Once again, 10 of these 20 outcomes include NEITHER Sergio NOR Rani. Once again, these 10 outcomes include group likes {T, U, V}, {T, V, X} and {U, V, W}

So, there are 10 outcomes that include NEITHER Rani NOR Sergio, and we have counted these outcomes in case 1 AND case 2.

To account for this duplication, we must subtract 10 from your solution to get 40 - 10 = 30

So, the correct answer is E.

## Hi Brent,

I still didn't get what you said about "Notice the 10 of these 20 outcomes do not include Rani" in Case 1 and "Notice the 10 of these 20 outcomes do not include Sergio" in Case.

Should this be other way i.e Rani and Sergio be replaced in your statements.

After that also, I didn't quite get the logic as what should be included and what (DUPLICATES) should be subtracted (REMOVED).

Does the question say that the outcome should contain atleast Rani or Sergio in the committe ?? I guess that is why the duplicates of NEITHER has been removed.

## Your question: "Does the

Your question: "Does the question say that the outcome should contain at least Rani or Sergio in the committee?? I guess that is why the duplicates of NEITHER has been removed."

No, we can have committees that contain neither Rani nor Sergio. My point is that we have double-counted all of the committee that contain neither Rani nor Sergio.

I have edited my response above to reflect this.

## Hi Brent i tried a different

## Hi prakap,

Hi prakap,

What you are describing is ONE possible case that needs to be considered to solve this question WITHOUT applying the restriction strategy (as described in the video).

If we don't apply the restriction strategy, then we must recognize that there are TWO different cases that satisfy the condition that R and S cannot both be on the same committee.

CASE 1: One of either R or S is on the committee, and 2 other committee members are selected from the other 5 people (T, U, V, W, and X)

CASE 2: Neither R nor S are on the committee, so all 3 committee members are selected from the other 5 people (T, U, V, W, and X)

Let's determine the number of outcomes for each case.

CASE 1:

- select 1 person from R and S. This can be accomplished in 2 ways

- select 2 people from T, U, V, W, and X. This can be accomplished in 5C2 ways (= 10 ways)

- TOTAL number of committees = (2)(10) = 20

CASE 2:

- select 0 people from R and S. This can be accomplished in 1 way

- select all 3 people from T, U, V, W, and X. This can be accomplished in 5C3 ways (= 10 ways)

- TOTAL number of committees = (1)(10) = 10

Answer = CASE 1 + CASE 2

= 20 + 10

= 30

Answer: E

Does that help?

Cheers,

Brent

## Brent,

Here is a similar question where Rani and Sergio aren't mentioned. However, I am getting stumped with choosing married couple. Let me explain, below is the question.

A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A) 16

B) 24

C) 26

D) 30

E) 32

Solution:

Total number of committees possible = 8C3 = 56

Selecting a committee of 3 with a married couple A, B and another person C:

Number of ways selecting married couple = 4 ways <- DON'T UNDERSTAND HOW?!

Number of ways of selecting the remaining = 6c1 = 6

Committee members without married couple = Total members - Married couple members

= 56 - 6 * 4 = 32

I am wondering how 4 came into the picture. Among 8 members isn't there 8c2 ways of people getting married? Isn't this similar to the handshake problem studied earlier?

Please help!

## Your question: 4 ways <- DON

Your question: 4 ways <- DON'T UNDERSTAND HOW?!

To ensure that no couples are on the committee together, we'll select ONE person from each of 3 couples.

So, let's first identify 3 couples.

There are 4 couples altogether.

So, we can choose 3 couples in 4C3 ways (= 4 ways)

Your question: I am wondering how 4 came into the picture. Among 8 members isn't there 8c2 ways of people getting married?

This isn't similar to the handshake problem, because the couples are ALREADY married.

8C2 suggests that the 8 people are all single, and we're going to pair them up to get married.

------HERE'S MY APPROACH------------------------

Basically, we need to select 1 person from each of 3 couples. This will ensure no couples are on the committee together.

So, one way to do this is to first identify which 3 couples we'll select people from.

There are 4 couples altogether.

So, we can choose 3 couples in 4C3 ways (= 4 ways)

Now that we've identified 3 couples, we'll choose 1 person from each couple.

We can choose one person from the first couple in 2 ways.

We can choose one person from the second couple in 2 ways.

We can choose one person from the third couple in 2 ways.

So, the total number of outcomes = 4 x 2 x 2 x 2 = 32

Here's my step-by-step solution: https://gmatclub.com/forum/a-committee-of-3-people-is-to-be-chosen-from-...

Does that help?

Cheers,

Brent

## Excellent answer! I never

## sir i did in a little

case 1 rani is on team= 10 ways

case 2 sergio is on team= 10 ways

case 3 none of them is on team= 10 ways

## That works! Cheers, Brent

That works!

Cheers,

Brent

## https://gmatclub.com/forum

sir please explain :)

## Question link: https:/

Question link: https://gmatclub.com/forum/how-many-different-ways-can-a-group-of-6-peop...

I don't like this question, because it's unclear whether the teams are considered identical.

For example, if the people are A, B, C, D, E and F, then we could have:

Team 1 = A & B, Team 2 = C & D, and Team 3 = E & F

Is this outcome different from: Team 1 = C & D, Team 2 = A & B, and Team 3 = E & F?

It's hard to say.

Given this ambiguity, I don't think this is a GMAT-worthy question.

Cheers,

Brent

## https://gmatclub.com/forum

sir in this i used Mississippi rule

5 in favour let them be aaaaa

and 2 against let them be bb

so we can arange them as 7!/5!*2!

is this correct?

## Question link: https:/

Question link: https://gmatclub.com/forum/members-of-a-student-parliament-took-a-vote-o...

Your solution is perfectly valid. Nice work!

Cheers,

Brent

## https://gmatclub.com/forum

sir in this i calculate manually

but while reading others answers i cant understand why we are dividing by 2 why half will be on left and half will be on right?

## Question link: https:/

Question link: https://gmatclub.com/forum/seven-children-a-b-c-d-e-f-and-g-are-going-to...

If we first ignore the rule about child C sitting to the right of A & B, then we get a total of 1440 possible outcomes.

Of those 1440 possible outcomes, half will have child C sitting to the RIGHT of A & B, and half will have child C sitting to the LEFT of A & B

Here's why: for every arrangement in which child C sitting to the RIGHT of A & B, we can take child a switch places with A & B to create an outcome in which child C is sitting to the LEFT of A & B

For example if one arrangement is: fgABedC, then we can create the arrangement fgCedAB by having C switch places with A & B.

Given that HALF of the 1440 arrangements will have child C sitting to the RIGHT of A & B and HALF will have child C sitting to the LEFT of A & B, we must divide 1440 by 2.

Does that help?

Cheers,

Brent

## https://gmatclub.com/forum/a

please explain

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/a-shipping-company-has-four-empty-trucks-that...

Cheers,

Brent

## https://gmatclub.com/forum/a

please explain

## Question link: https:/

Question link: https://gmatclub.com/forum/a-plant-manager-must-assign-10-new-workers-to...

As you can see from the discussion in the thread, it's a poorly-worded question.

In my opinion, it's not a GMAT-worthy question.

Skip it.

Cheers,

Brent

## Hey Brent,

Can i solve it by taking Rani and Sergio to be one entity and then select 3 persons by 6C3. And then take 7C3 by taking all combinations into account.

I got 70-40=30.

## Sorry, but I don't understand

Sorry, but I don't understand your reasoning.

If you combine Rani and Sergio to be ONE entity and then select 3 entities (in 6C3) ways, then you COULD be selecting 4 people (if you select Uma, Vivek, and the Rani-Sergio entity).

Also, where do the numbers 70 and 40 come from?

6C3 = 20, and 7C3 = 35

Cheers,

Brent

## I thought that by taking Rani

I keep getting confused when I calculate combinations using the shortcut method. So, instead of dividing by 3!, I divided by 3. Sheer luck that I ended up with the right answer with the wrong procedure. :)

## Your strategy can still work.

Your strategy can still work.

Once we're transformed Rani and Sergio into ONE entity, then we can create a 3-person committee by selecting 1 more person from the remaining 5 people.

We can do this in 5 ways.

So, there are 5 ways to have Rani and Sergio on the SAME committee (and thus break the restriction).

If we IGNORE the restriction, we can create a 3-person committee in 7C3 ways (=35 ways)

So, total number of allowable arrangements = 35 - 5 = 30

Cheers,

Brent

## Oh great! I'm happy that

Thanks.

## Add a comment