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Ben and Ann are among 7 contestants from which 4

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Ben and Ann are among 7 contestants from which 4 [#permalink] New post 17 Mar 2008, 06:47
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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




Please explain your answer
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 07:18
Choose 4 from 5 persons (- Ben,Ann) 4C5 =5
What is the OA?
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 07:51
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 ;-)
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 07:56
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




Please explain your answer

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

5c4 = 5

a
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 12:50
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).
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 13:52
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
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Re: PS: Combination [#permalink] New post 17 Mar 2008, 20:49
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
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Re: PS: Combination [#permalink] New post 18 Mar 2008, 05:24
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
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Re: PS: Combination [#permalink] New post 18 Mar 2008, 07:27
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.
Re: PS: Combination   [#permalink] 18 Mar 2008, 07:27
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