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

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

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New post 16 Oct 2016, 05:54
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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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Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten  [#permalink]

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New post 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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New post 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

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

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

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New post 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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New post 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!!!
_________________

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GMATinsight
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e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

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

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

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New post 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.

Answer: D
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Re: Bert and Ernie are among 12 tenants on Sesame Street, from which 9 ten &nbs [#permalink] 07 Sep 2018, 16:06
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