Question: Committees without R and S

Comment on Committees without R and S

Why are you making 4 person committees when it says 3 person committees in the question?
greenlight-admin's picture

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

How many possible questions on Counting should be in the GRE test?
greenlight-admin's picture

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

What should the real answer be with a three person committee? I am getting 25??

greenlight-admin's picture

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 second stage( when we are breaking the rule)? Thank you so much!
greenlight-admin's picture

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 Rule and use that in the future, but I tried a different approach and I was wondering if you could tell me why it's wrong.

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?
greenlight-admin's picture

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 the rule was we use the fundamental counting theorem to find the number of possible arrangements (permutations) and then if it's a combination, we divide by the number of slots or "r!" to eliminate double counting of arrangements that are the same given that order doesn't matter. In the note example though, we divide by less than the number of slots. It seems the rule would be we don't divide by the number of "r!" if one of the slots is 1? We divide by (r - predetermined choices)! Correct?

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)!

greenlight-admin's picture

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 be by (r - choices that are 1)!
greenlight-admin's picture

That would work in this case.
Alternatively, the applying the Restrictions Rule would be the best option.

Thanks a lot. I appreciate it.

For the breaking restrictions segment, why is it 5 (does that indicate FCP of 1x1x5?) and not 2x1x5 (indicating that the first position could be filled by either Sergio or Rani, and second position filled by the remaining one out of the two?) Thanks!
greenlight-admin's picture

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?

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