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Ben and Ann are among 7 contestants from which 4

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by [email protected] » Wed Dec 11, 2019 9:09 pm
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.

Final Answer: A

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by Brent@GMATPrepNow » Thu Dec 12, 2019 5:45 am
BTGmoderatorDC 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
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

Answer: A

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by Scott@TargetTestPrep » Thu Dec 12, 2019 6:34 pm
BTGmoderatorDC 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


OA A

Source: Official Guide
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.

Answer: A

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by deloitte247 » Thu Dec 19, 2019 9:40 pm
There are 7 contestants.
4 semi-finalists are to be selected.
Four different possible selection with neither Ann nor Bens
$$=\left(7-2\right)C_4$$
$$=5_{C_4}=\ \frac{5!}{4!}=\frac{5\cdot4!}{4!}=5$$
Answer = Option A
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