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Rani and Segio

by jain2016 » Sat Mar 05, 2016 11:52 am
A 3-person committee must be selected from seven people: Rani, Sergio, Vivek, Takumi, Uma, Walter, Xavier. If Rani and Sergio can not be on the same committee, how many different committees are possible?

A) 10

B) 15

C) 20

D) 25

E) 30

Hi Expert ,

Please check and advise.

# of ways to select 3 out of seven members is 7C3 = 35

# of ways to select Rani and Sergio on the same committee is 2X1X5 = 10 ( because for the 1st place we have either Rani or Sergio so we can fill in 2 ways, then for the 2nd ways we have to select between Rani and Sergio so this place can be filled by 1 way and for the 3rd place we have remaining 5 members, so this place can be filled by 5 ways.)

so # of different committees in which Rani and Sergio can be part of same team = 35 - 10 = 25

But OA is E

Please correct me.

Many thanks in advance.

SJ
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by Brent@GMATPrepNow » Sat Mar 05, 2016 1:23 pm
jain2016 wrote: # of ways to select Rani and Sergio on the same committee is 2X1X5 = 10 ( because for the 1st place we have either Rani or Sergio so we can fill in 2 ways, then for the 2nd ways we have to select between Rani and Sergio so this place can be filled by 1 way and for the 3rd place we have remaining 5 members, so this place can be filled by 5 ways.)
The problem is with this calculation.
The order in which we select the committee members does not matter. However, you are saying that the first 2 selections must include Rani and Sergio.
To make it clearer, what if we were to create that 2-person committee that MUST include both Rani and Sergio. In how many ways can we do this? Well, there's only 1 such committee. However, if we use your (incorrect) approach, we would say that we can select one of TWO people (2 options) for the first position, and then select the remaining person for the second position. So, the total number of 2-person committees = (2)(1) = 2
Of course this isn't correct.

If we're going to use the slot method (aka stages approach), we must ask, "Is the outcome of one stage different from the outcome of another stage?" If they are different, we can continue with the slot/stages approach. If they are the same, we must use a different approach.
In this question, selecting someone first and selecting someone second and selecting someone third all have the same outcome - each selection gets to be on the committee. As such, we can't use the slot/stages approach.

For more on this idea, see this free video: https://www.gmatprepnow.com/module/gmat ... /video/788

To watch the complete video answer to this question, see: https://www.gmatprepnow.com/module/gmat ... /video/797

Cheers,
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by [email protected] » Sat Mar 05, 2016 5:21 pm
Hi jain2016,

You are correct that there are 35 possible groups of 3 people - since we're dealing with the total combinations of people, and the order of the people does NOT matter, we use the Combination Formula . Since Rani and Sergio CAN'T be on the committee together, we have to remove any committee that they both appear on. That's actually a short list:

R and S and V
R and S and T
R and S and U
R and S and W
R and S and X

5 of the 35 committees have to be removed. 35 - 5 = 30

Final Answer: E

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by GMATGuruNY » Sat Mar 05, 2016 11:36 pm
jain2016 wrote:A 3-person committee must be selected from seven people: Rani, Sergio, Vivek, Takumi, Uma, Walter, Xavier. If Rani and Sergio can not be on the same committee, how many different committees are possible?

A) 10

B) 15

C) 20

D) 25

E) 30

Hi Expert ,

Please check and advise.

# of ways to select 3 out of seven members is 7C3 = 35

# of ways to select Rani and Sergio on the same committee is 2X1X5 = 10 ( because for the 1st place we have either Rani or Sergio so we can fill in 2 ways, then for the 2nd ways we have to select between Rani and Sergio so this place can be filled by 1 way and for the 3rd place we have remaining 5 members, so this place can be filled by 5 ways.)
The portion in red is incorrect.
A BAD committee consists of Rani, Sergio and one other person.
Thus, to form a BAD committee, we must select ONE person to be combined with Rani and Sergio.
From the 5 other people, the number of ways to select one to be combined with Rani and Sergio = 5.

Subtracting these 5 bad committees from the 35 possible committees, we get:
35-5 = 30.

The correct answer is E.
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by nidhiranpara » Wed Mar 30, 2016 9:30 am
I have new approach to this question.
There are total 7 people and question tells us that Rani and Sergio can not be on the same committee.
so first step, we have to choose 3 out of 7 with above condition.so for 1st place we can either choose Rani or Sergio this we can write as 2C1=2.
Now we have 5 people left for 2 places so we can choose that by 5C2=10
so total is 2C1 X 5C2 = 2 X 10 = 20
this is one possible choice other possible choice is we neither choose Rani nor Sergio. so we left with 5 people to fill up 3 places. this can be done by 5C3=10
so final ans is
(2C1 X 5C2) + (5C3)= 20 + 10 = 30 :)
I hope my this approach is right!
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by MartyMurray » Wed Mar 30, 2016 1:49 pm
nidhiranpara wrote:I hope my this approach is right!
You nailed it.
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