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# Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten

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Math Expert
Joined: 02 Sep 2009
Posts: 64174
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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16 Oct 2016, 05:54
2
3
00:00

Difficulty:

25% (medium)

Question Stats:

76% (01:31) correct 24% (02:08) wrong based on 162 sessions

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Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

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Joined: 18 Jun 2016
Posts: 85
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 700 Q49 V36
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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16 Oct 2016, 06:07
1
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Neither Bert or Ernie , so list will have 10 people
Question now becomes choose 9 out of 10

10! / 1!*9!
= 10

there are 10 ways to do so.
IMO D
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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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03 Sep 2018, 22:47
1
stne wrote:
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Hi chetan2u,

Why am I not getting the answer the other way, finding total ways to select 9 out of 12 ( 12C9) then subtracting all cases where Bert and Ernie are together .

12C9= $$\frac{12!}{3!9!}$$ =220

After Choosing Bert and Ernie, now we have 7 places left and 10 people to choose from , hence 10C7= 120

220 -120 =100

What am I missing here ? Thank you.

stne

You have to subtract all cases in which any of the Bert and Ernie are part of the team

i.e. Correct Answer = Total Possible teams of 9 - Teams of 9 with Earnie among them - Team of 9 with Bert among them + Team of 9 with Earnie and Burt both among them

i.e. 12C9 - 11C8 - 11C8 + 10C7 = 220 - 165 - 165 + 120 = 10

I hope this helps!!!
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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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03 Sep 2018, 22:49
1
stne wrote:
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Hi chetan2u,

Why am I not getting the answer the other way, finding total ways to select 9 out of 12 ( 12C9) then subtracting all cases where Bert and Ernie are together .

12C9= $$\frac{12!}{3!9!}$$ =220

After Choosing Bert and Ernie, now we have 7 places left and 10 people to choose from , hence 10C7= 120

220 -120 =100

What am I missing here ? Thank you.

Alternatively you could also do,

Favorable cases = 12C9 - Team with E and B both together - Team of 9 with E but not with B - Team of 9 with B but not E

Favorable cases = 12C9 - 10C7 - 10C8 - 10C8 = 220 - 120 - 45 - 45 = 10

I hope this helps!!!
_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes l Classroom Coaching l On-demand Quant course
Check website for most affordable Quant on-Demand course 2000+ Qns (with Video explanations)
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3992
Location: India
GMAT: QUANT EXPERT
Schools: IIM (A)
GMAT 1: 750 Q51 V41
WE: Education (Education)
Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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16 Oct 2016, 06:11
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Total ways of selecting 9 out of 10 leaving bert and ernie = 10C9 = 10

_________________
Prosper!!!
GMATinsight .............(Bhoopendra Singh and Dr.Sushma Jha)
e-mail: info@GMATinsight.com l Call : +91-9999687183 / 9891333772
Online One-on-One Skype based classes l Classroom Coaching l On-demand Quant course
Check website for most affordable Quant on-Demand course 2000+ Qns (with Video explanations)
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
ACCESS FREE GMAT TESTS HERE:22 FREE (FULL LENGTH) GMAT CATs LINK COLLECTION
VP
Joined: 27 May 2012
Posts: 1029
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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03 Sep 2018, 09:44
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Hi chetan2u,

Why am I not getting the answer the other way, finding total ways to select 9 out of 12 ( 12C9) then subtracting all cases where Bert and Ernie are together .

12C9= $$\frac{12!}{3!9!}$$ =220

After Choosing Bert and Ernie, now we have 7 places left and 10 people to choose from , hence 10C7= 120

220 -120 =100

What am I missing here ? Thank you.
_________________
- Stne
VP
Joined: 27 May 2012
Posts: 1029
Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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04 Sep 2018, 00:37
GMATinsight wrote:
stne wrote:
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

Hi chetan2u,

Why am I not getting the answer the other way, finding total ways to select 9 out of 12 ( 12C9) then subtracting all cases where Bert and Ernie are together .

12C9= $$\frac{12!}{3!9!}$$ =220

After Choosing Bert and Ernie, now we have 7 places left and 10 people to choose from , hence 10C7= 120

220 -120 =100

What am I missing here ? Thank you.

Alternatively you could also do,

Favorable cases = 12C9 - Team with E and B both together - Team of 9 with E but not with B - Team of 9 with B but not E

Favorable cases = 12C9 - 10C7 - 10C8 - 10C8 = 220 - 120 - 45 - 45 = 10

I hope this helps!!!

Thank you, this is clear now .
_________________
- Stne
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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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07 Sep 2018, 16:06
Bunuel wrote:
Bert and Ernie are among 12 tenants on Sesame Street, from which 9 tenants are to be selected for the neighborhood watch. Of the different possible selections, how many contain neither Bert nor Ernie?

A. 1
B. 3
C. 9
D. 10
E. 12

The number of selections that contain neither Bert nor Ernie is 10C9 = 10.

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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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18 Feb 2020, 13:03
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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten   [#permalink] 18 Feb 2020, 13:03