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# Ben and Ann are among 7 contestants from which 4 semifinalists are to

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Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Let the remaining contestants be A,B,C,D, and E.

They are listed as follows
ABCD | ABCE | BCDE | ACDE | ABDE

Therefore, there are $$C_4^{5} = 5$$(Option A) possible combinations for the semifinalists.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
Bunuel, could you please explain? Thanks.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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gmatt1476 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

PS07602.01

Given:Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected.

Asked: Of the different possible selections, how many contain neither Ben nor Ann?

There are 5 other participants out of which 4 are to be selected

Number of ways = 5C4 = 5

IMO A

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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Hi All,

We're told that Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. We're asked of the different possible selections, how many contain NEITHER Ben nor Ann. Since we're selecting 'groups' of people, this is ultimately a Combination Formula question.

IF we had NO restrictions, then with 7 contestants, the number of 'groups of four' would be 7c4 = 7!/4!3! = (7)(6)(5)/(3)(2)(1) = 35 possible groups of four

However, we're interested in groups that contain NEITHER Ben nor Ann, which means that we have just 5 people to choose from. 5c4 = 5!/4!/1! = (5)/(1) = 5 groups of four.

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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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gmatt1476 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

PS07602.01

Let Ann, Bob, C, D, E, F, G be the 7 contestants.

To ensure that neither Ben nor Ann are among the four semifinalists, let's remove them from the list of contestants.

So, we can select the four semifinalists from {C, D, E, F, G}

Since the order in which we select the 4 semifinalists does not matter, we can use COMBINATIONS.
We can select 4 semifinalists from 5 contestants in 5C4 ways.
5C4 = 5

Cheers,
Brent
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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gmatt1476 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

PS07602.01

If neither Ben nor Ann is among the semifinalists, then there are only 5 contestants vying for the four slots. Thus, the number of ways to select the semifinalists when neither Ben nor Ann is selected is 5C4 = 5.

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Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
It might be very silly but I have done it like 7C4- 2C2*5C2.
Total- ways of choosing Ben and Ann and * ways of choosing the remaining. What is the conceptual mistake I am making?
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Hi Anki3188,

Since the question is asking for the number of groups that do NOT include Ben AND do NOT include Ann, then your calculation would need to remove any group that includes Ben OR Ann OR both of them. Your calculation appears to only remove the groups that include both Ben and Ann.

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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
chetan2u Bunuel I know I am coming up with the wrong answer, but can you help me understand what is wrong with the below approach -
[Total number of ways to select 4 candidates out of 7] - [number of ways Ben and Ann can be selected in the semifinalists]

Total number of ways to select 4 candidates out of 7 = 7C4 = 35

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5 ways = 5
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5
ben and ann together as semifinalists = 1 way
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways

=> 35 - 11 = 24 ways
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Argp wrote:
chetan2u Bunuel I know I am coming up with the wrong answer, but can you help me understand what is wrong with the below approach -
[Total number of ways to select 4 candidates out of 7] - [number of ways Ben and Ann can be selected in the semifinalists]

Total number of ways to select 4 candidates out of 7 = 7C4 = 35

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5 ways = 5
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5
ben and ann together as semifinalists = 1 way
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways

=> 35 - 11 = 24 ways

You have gone wrong in your calculations

number of ways Ben is selected in the semifinalists (except Ann) = B_ = 1x5C3 ways = 10....You have to select 3 from remaining 5.
number of ways Ann is selected in the semifinalists (except Ben) = A_ = 1x 5 ways = 5...10 ways as above
ben and ann together as semifinalists = 1 way .......2C2*5C2=10
number of ways Ben and Ann can be selected in the semifinalists => 5+ 5 + 1 = 11 ways.......10+10+10=30 ways

Total = 35-30=5

Anki3188 , you have gone wrong in not taking cases where only one of them is selected. The answer would be correct if the question were to find ways in which both are not there.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
I am trying to do this using the counting method and would appreciate some help where I'm going wrong. I got the correct answer using formulas but cant seem to with the counting method.

So basically

Total ways of selecting 4 of 7 will give us all the ways - 7x6x5x4

And subtracting the following should give us the desired answer

1. Where Ben is a semifinalist but Ann is not - 1 x 5 x 4 x 3 ----- (a)
2. Where Ann is a semifinalist but Ben is not - 1 x 5 x 4 x 3 ----- (b)

T -(a+b) = 840 -(60+60) = 740 --- ???
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
I am trying to do this using the counting method and would appreciate some help where I'm going wrong. I got the correct answer using formulas but cant seem to with the counting method.

So basically

Total ways of selecting 4 of 7 will give us all the ways - 7x6x5x4

And subtracting the following should give us the desired answer

1. Where Ben is a semifinalist but Ann is not - 1 x 5 x 4 x 3 ----- (a)
2. Where Ann is a semifinalist but Ben is not - 1 x 5 x 4 x 3 ----- (b)

T -(a+b) = 840 -(60+60) = 740 --- ???

It is selection and you are taking arrangements.

So T = 7C4 = 35

a) B is one, so choose remaining 3 from available 5( less A and B) = 5C3 = 10
b) A and n(B): Similarly 10 ways for A being a part and B not being a part.
c) both (A and B): Also add ways where both A and B are part, so choose 2 out of remaining 5 = 5C2 = 10
Our answer = 35-10-10-10 = 5
A

There are various combinations in which above 3 could be calculated.
It could be (T-(A as one)-(B there but not A))=35-6C3-5C3=35-20-10=5
OR T-(A)-(B)+both(A and B)=35-20-20+10=5
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
BrentGMATPrepNow wrote:
gmatt1476 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

PS07602.01

Let Ann, Bob, C, D, E, F, G be the 7 contestants.

To ensure that neither Ben nor Ann are among the four semifinalists, let's remove them from the list of contestants.

So, we can select the four semifinalists from {C, D, E, F, G}

Since the order in which we select the 4 semifinalists does not matter, we can use COMBINATIONS.
We can select 4 semifinalists from 5 contestants in 5C4 ways.
5C4 = 5

Cheers,
Brent

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Kimberly77 wrote:

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30

If you want to go that route you need to follow the property that says:
Number of ways to follow the restriction = (number of ways to ignore the restriction) - (number of ways to break the restriction)

The restriction here is that neither Ben nor Ann can be in the group of 4 semi finalists.

So we get:
Number of outcomes in which neither Ben nor Ann are in the group of 4 semi finalists = (number of 4-person groups chosen from from all seven semi-finalists) - (number of ways to break the restriction)

Here, your calculation of 5C4 = 5 does not give us the total number of ways to BREAK the restriction.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:

Hi BrentGMATPrepNow, not sure why if I calculate all contestants first then minus Ben and Ann is not correct?

7C4 = 35
5C4 = 5
35-5 = 30

If you want to go that route you need to follow the property that says:
Number of ways to follow the restriction = (number of ways to ignore the restriction) - (number of ways to break the restriction)

The restriction here is that neither Ben nor Ann can be in the group of 4 semi finalists.

So we get:
Number of outcomes in which neither Ben nor Ann are in the group of 4 semi finalists = (number of 4-person groups chosen from from all seven semi-finalists) - (number of ways to break the restriction)

Here, your calculation of 5C4 = 5 does not give us the total number of ways to BREAK the restriction.

Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
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Kimberly77 wrote:
Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?

The restriction is that the 4-person selection CANNOT contain Ben or Ann.
Since 5C4 is the total number of 4-person selections that do NOT contain Ben or Ann, then you aren't breaking the restriction; you're obeying the restriction.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Thanks BrentGMATPrepNow, in 5C4 = 5 which I have already removed Ben and Ann from 7 therefore 5 contestants left, then select 4 out of 5 semi-finalists. So not sure why this still doesn't include the total number of ways to BREAK the restriction? Could we not remove both Ben and Ann at the same time since question asked neither Ben nor Ann are to be in semi-finalists?

The restriction is that the 4-person selection CANNOT contain Ben or Ann.
Since 5C4 is the total number of 4-person selections that do NOT contain Ben or Ann, then you aren't breaking the restriction; you're obeying the restriction.

Thanks BrentGMATPrepNow, think I get it now.
So we need to calculate 3 cases of no Ben 5C3, no Ann 5C3, no Ben & Ann 2C2 * 5C2. Total 35-30 = 5

Mistake I made here was I interpret neither Ben nor Ann as both not on there. Therefore only calculate it once. How will I know to interpret neither Ben nor Ann as either or? Thanks Brent
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