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Ben and Ann are among 7 contestants from which 4

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by BTGModeratorVI » Fri May 29, 2020 6:47 am

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Answers

A

B

C

D

E

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Difficulty—

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

Answer: A
Source: Official guide
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Re: Ben and Ann are among 7 contestants from which 4

by Brent@GMATPrepNow » Fri May 29, 2020 8:51 am
BTGModeratorVI wrote: ↑
Fri May 29, 2020 6:47 am
 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

Answer: A
Source: Official guide
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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: Ben and Ann are among 7 contestants from which 4

by ezmodest » Sat May 30, 2020 9:37 am
This is a nice combinatorics problem with a little twist. The question is asking for every combination possible that doesn't include Ben OR Ann & Ben & Ann. Rephrasing like this simplifies the problem, because we now know that out of the 7 contestants we don't want Ben or Ann to even be considered as semifinalists, so we can effectively remove them.

This leaves us with the following:
7 (starting contestants) - 2 (Ben and Ann) = 5 (remaining contestants)

Since we have 5 contestants and 4 spaces AND order doesn't matter (they are all semifinalists, aka in what order they become semifinalist doesn't affect the outcome), we can say the following:

5 choose 4 = $$\frac{5!}{\text{4!1!}}$$ = 5

Answer: A
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