It is currently 11 Dec 2017, 14:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Ben and Ann are among 7 contestants from which 4

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
SVP
Joined: 21 Jul 2006
Posts: 1509

Kudos [?]: 1070 [0], given: 1

Ben and Ann are among 7 contestants from which 4 [#permalink]

### Show Tags

17 Mar 2008, 07:47
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

a) 5

b) 6

c) 7

d) 14

e) 21

Kudos [?]: 1070 [0], given: 1

Manager
Joined: 02 Mar 2008
Posts: 205

Kudos [?]: 53 [0], given: 1

Concentration: Finance, Strategy
Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 08:18
Choose 4 from 5 persons (- Ben,Ann) 4C5 =5
What is the OA?

Kudos [?]: 53 [0], given: 1

Senior Manager
Joined: 15 Aug 2007
Posts: 252

Kudos [?]: 73 [0], given: 0

Schools: Chicago Booth
Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 08:51
AlbertNTN wrote:
Choose 4 from 5 persons (- Ben,Ann) 4C5 =5
What is the OA?

A-5

But the correct representation is 5C4. You can't chose 5 from 4

Kudos [?]: 73 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1087 [0], given: 4

Location: New York City
Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 08:56
tarek99 wrote:
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain neither Ben nor Ann?

a) 5

b) 6

c) 7

d) 14

e) 21

the original selection is 7C4
u take out the 2 ppl that u dont want --> 7-2 = 5

5c4 = 5

a
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1087 [0], given: 4

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 536 [0], given: 0

Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 13:50
I get 5 as well.

7!/4!3! = 35 number of ways w/ no constraints.

Now we have VWXYZAB

There are 5!/4!1! ways to arrange VWXYZ into groups of 4.

So its just 5.

Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

Kudos [?]: 536 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1087 [0], given: 4

Location: New York City
Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 14:52
GMATBLACKBELT wrote:
I get 5 as well.

7!/4!3! = 35 number of ways w/ no constraints.

Now we have VWXYZAB

There are 5!/4!1! ways to arrange VWXYZ into groups of 4.

So its just 5.

Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain both Ben and Ann?

if B and A are in it, then we only have to choose 2 of 5 people.

1*1*5C2 / 7c4
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1087 [0], given: 4

SVP
Joined: 29 Aug 2007
Posts: 2470

Kudos [?]: 867 [0], given: 19

Re: PS: Combination [#permalink]

### Show Tags

17 Mar 2008, 21:49
GMATBLACKBELT wrote:
Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

make sure ts 10 not 30. you probably did as under:
total = 7c4 = 35
without A and B = 5
so with A and B = 35 - 5 = 30
but thats wrong because 30 includes where A is but not B and B is in but not A. so the correct one is:

5c2 = 5x4x3!/3!x2 = 10
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Kudos [?]: 867 [0], given: 19

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1087 [0], given: 4

Location: New York City
Re: PS: Combination [#permalink]

### Show Tags

18 Mar 2008, 06:24
bmwhype2 wrote:
GMATBLACKBELT wrote:
I get 5 as well.

7!/4!3! = 35 number of ways w/ no constraints.

Now we have VWXYZAB

There are 5!/4!1! ways to arrange VWXYZ into groups of 4.

So its just 5.

Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain both Ben and Ann?

if B and A are in it, then we only have to choose 2 of 5 people.

1*1*5C2 / 7c4

whoops. gave u the probability. just take the numerator
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1087 [0], given: 4

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 536 [0], given: 0

Re: PS: Combination [#permalink]

### Show Tags

18 Mar 2008, 08:27
GMAT TIGER wrote:
GMATBLACKBELT wrote:
Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

make sure ts 10 not 30. you probably did as under:
total = 7c4 = 35
without A and B = 5
so with A and B = 35 - 5 = 30
but thats wrong because 30 includes where A is but not B and B is in but not A. so the correct one is:

5c2 = 5x4x3!/3!x2 = 10

Sorry I shouldve been more clear. I meant all of the ways w/ A and B, also Only A and Only B.

Kudos [?]: 536 [0], given: 0

Re: PS: Combination   [#permalink] 18 Mar 2008, 08:27
Display posts from previous: Sort by

# Ben and Ann are among 7 contestants from which 4

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.