# Question: Committees without R&S

## Comment on Committees without R&S

### How can we solve this without

How can we solve this without using restrictions rule?

### 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

I tried a different approach (guessed) and came up with the right answer. I tried combining R & S as one so that I'd have 5 people to choose from (7-2 = 5) to make a committee of 3. _ _ | _ The first 2 slots were the 5c3 and then the final slot was 3c1. I was wondering, since i guessed, if this was a valid 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

On listing the committees (at around 26 secs into the video), why are there 4 people being listed?

### Good catch! It should be

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

### Hi Brent,

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.

### 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,

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 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 Brent i tried a different approach, we take R & S as one and select from them in 2C1=2 ways and the remaining 5 in 5C2=10 ways...after this I'm lost , so please help, is this even a valid approach??

### 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

Does that help?

Cheers,
Brent

### 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?

### 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 thought about the problem the way you explained Thanks!!

### sir i did in a little

sir i did in a little different way
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

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

https://gmatclub.com/forum/how-many-different-ways-can-a-group-of-6-people-be-divided-into-3-team-219479.html

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

https://gmatclub.com/forum/members-of-a-student-parliament-took-a-vote-on-a-proposition-for-a-new-234699.html
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?

Your solution is perfectly valid. Nice work!

Cheers,
Brent

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

https://gmatclub.com/forum/seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194099.html
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?

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

https://gmatclub.com/forum/a-plant-manager-must-assign-10-new-workers-to-one-of-five-shifts-she-204057.html

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,

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 thought that by taking Rani and Sergio as one entity I can calculate the number of ways in which both of them will be in the committee together. Then I subtracted it from the total number of ways a team can be selected from the 7 members.

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.

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

Oh great! I'm happy that atleast my line of thought was correct to some point.
Thanks.

### Hi Brent, what would be the

Hi Brent, what would be the solution if the committee was for 4 people or more instead of three. How would we do the final part of the calculation that is currently easy.

### Let's say we're choosing 4

Let's say we're choosing 4 people to be on the committee.

If we IGNORE the restriction, we simply select any 4 people from 7 people to be on the committee.
This can be achieved in 7C4 ways (= 35 ways)

To determine the number of outcomes that BREAK the restriction, place R and S on the committee, and then choose 2 more people to be on the committee.
We can choose 2 people from the remaining 5 people in 5C2 ways (= 10 ways)

So the number of ways to FOLLOW the restriction = 35 - 10 = 25

Cheers, Brent

### Hi Brent

Hi Brent
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

In this question : why 5c3 and why not 10c2. Can you pls expalin. Thanks

In my solution, I am choosing 3 couples out of the 5 couples (which can be accomplished in 5C3 ways).
Then, from each of the 3 selected couples, I choose 1 person.
This ensures that we don't have any married people on the committee.

I'm not sure why we'd use 10C2. Can you elaborate on your solution?

### Rani sergio prob

Rani sergio prob
7c2 - 5 ways why 5 ways not understood, to break the restn.

### The Restriction here is that

The Restriction here is that Rani and Sergio cannot be on the same committee.
To BREAK this restriction, Rani and Sergi must be on the same committee.
To determine the number of different committees that include both Rani and Sergio, we'll first place those two people on the committee.
Then we'll choose 1 more person to join them.
Since there are 5 people remaining, we can choose the third person in 5 ways.

If you're not convinced, here are the five possible committees that include both Rani and Sergio:
- RST
- RSU
- RSV
- RSW
- RSX

Does that help?

### Hi Brent, I calculate it as 3

Hi Brent, I calculate it as 3 people to form committe out of 7 - R&S not on committe but is not getting the right answer? What is missing?

3 people to form committe out of 7 : 7C3 = 35
R&S not on committe : 5C3 = 10
35-10 = 25

### Notice that 7C3 is the total

Notice that 7C3 is the total number of ways to IGNORE the restriction.
So, we now we need to calculate the total number of ways to BREAK the restriction.

The 5C3 in your solution does not give us the total number of ways to BREAK the restriction.
What values do the 5 and the 3 and your solution represent?
If the 5 represents T, U, V, W and X, then 5C3 = the total number of ways to select three people from T, U, V, W and X.
Notice that the answer (10) does not tell us how many ways we can BREAK the restriction.

### Thanks Brent for the

Thanks Brent for the clarification. Still a bit confused.
In that case how's this question different to below one? Not sure why this question's calculation is not right with below one? Could you help? Thanks Brent for your great help always

### For the above video question,

For the above video question, the main condition is that R and S cannot BOTH be on the committee.
So for this question, R can be on the committee, and S can be on the committee.
They just can't BOTH be on the committee.
So, some acceptable outcomes are as follows:
STV
TUV
TVX
RTX
and so on.
Some outcomes that are NOT allowed include: RSX, and RSV.

For this linked question, A can never be on the committee, and B can never be on the committee.
So, if we let A,B,C,D,E and F be the 7 contestants, then some possible outcomes are:
CDE
CEF
CDF
Some outcomes that are NOT allowed include: ABD, ADE, and BDE.

### Great explanation thanks

Great explanation thanks Brent. For video question, we are calculating for *possible # of committes* without R&S at the same time. Therefore we removed R & S for calculation.

For Ben & Ann question, we are calculating for *different possible selections that contain neither Ben nor Ann*. Therefore we need to consider 3 cases: 1) no Ben in the selection 2) no Ann in the selection 3) no Ben & Ann on the selection.

Is my understanding correct? Thanks Brent.

PS. Now I see your way of calculating Ben & Ann (adhere to restriction) is the best and see why my calculation is wrong and time consuming. Thanks

### Ben and Ann question link:

Ben and Ann question link: https://gmatclub.com/forum/ben-and-ann-are-among-7-contestants-from-whic...
The "Ben and Ann question" can be solved in more than one way.
In order to exclude Ben and Ann from the committee, I just removed them, which allowed me to select four semi-finalists from the remaining 5 contestants.

Regarding your query, are you trying to solve the question using the strategy where:
Number of outcomes that SATISFY the conditions = (number of outcomes that IGNORE the condition) - (number of outcomes that BREAK the condition)?

If so then, when it comes to the number of outcomes that BREAK the condition, we must consider three cases:
1) Ann is selected, but Ben is not selected.
2) Ben is selected, but Ann is not selected.
3) Ben and Ann are both selected.

### Thanks Brent and yes that is

Thanks Brent and yes that is the strategy. Somehow I found that these 2 questions look so similar initially and yet so different as R&S is asking for # of possible committees but Ben and Ann is asking for # of possible selections with neither nor conditions. Very tircky.....

### Agreed. A lot of the trickier

Agreed. A lot of the trickier GMAT quantitative questions have a strong reading comprehension element.

### Noted thanks Brent.

Noted thanks Brent.