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

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Senior Manager
Joined: 04 Sep 2017
Posts: 291
Ben and Ann are among 7 contestants from which 4 semifinalists are to  [#permalink]

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21 Sep 2019, 04:22
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Difficulty:

35% (medium)

Question Stats:

66% (01:20) correct 34% (01:55) wrong based on 289 sessions

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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

PS07602.01
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Ben and Ann are among 7 contestants from which 4 semifinalists are to  [#permalink]

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21 Sep 2019, 04:42
2
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]

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21 Sep 2019, 05:14
3
1
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

total = 7 contestants and not having Ben & Ann ; we have 5 left out of which 4 semifinalists can be choosen in 5c4 ways ; 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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30 Sep 2019, 08:50
1
I find the wording of the problem statement to be confusing.

The way I interpreted it, the question asks for the total number of combinations of 4 out of 7 (7C4=35), and of those combinations, asks how many cannot contain Ben and Ann.

I went about solving it as 7C4 - 4C2 = 35 - 12 = 23.

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

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03 Oct 2019, 13:17
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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03 Oct 2019, 19:20
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

Posted from my mobile device
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to  [#permalink]

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22 Oct 2019, 09:17
2
1
icanhazmba wrote:
I find the wording of the problem statement to be confusing.

The way I interpreted it, the question asks for the total number of combinations of 4 out of 7 (7C4=35), and of those combinations, asks how many cannot contain Ben and Ann.

I went about solving it as 7C4 - 4C2 = 35 - 12 = 23.

You got the correct total number of combinations.
But, you need to exclude 3 other types of outcomes.

1) Choose Ben and 3 other contestants not including Ann: C1*5C3 = 10

2) Choose Ann and 3 other contestants not including Ben: C1*5C3 = 10

3) Choose Ann and Ben and 2 other contestants: 2C2*5C2 = 10

So the answer would be 7C4 - C1*5C3 - C1*5C3 - 2C2*5C2 = 35 - 10 - 10 - 10 = 5

Hope it's clear.
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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to  [#permalink]

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11 Dec 2019, 22:44
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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12 Dec 2019, 07:03
Top Contributor
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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13 Dec 2019, 14:24
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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Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to  [#permalink]

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06 Jan 2020, 14:04
MBA HOUSE KEY CONCEPT: Combinatorics (Combination)

7 contestants and 4 semifinalists

If we don’t want Benn and Ann actually we have 5 contestants and 4 semifinalists

n = 5 ; k = 4; no repetition; order doesn’t matter

Combination of 5 elements taken 4 by 4 = 5! / (5 - 4)! (4!) = 5

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

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06 Jan 2020, 23:43
Here, the required: how many contain neither Ben nor Ann?

Remove Ben and Ann from total contestants. Now total contestants stand 5.

As order of the contestants is not necessary, we can use combination.

From 5 , 4 can be selected in 5C4 = 5(A)
Re: Ben and Ann are among 7 contestants from which 4 semifinalists are to   [#permalink] 06 Jan 2020, 23:43
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