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# Ben and Ann are among 7 contestants from which 4

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Director
Joined: 03 Jul 2003
Posts: 656
Followers: 2

Kudos [?]: 22 [0], given: 0

Ben and Ann are among 7 contestants from which 4 [#permalink]  28 Jan 2004, 15:58
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19. 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
Manager
Joined: 26 Dec 2003
Posts: 227
Location: India
Followers: 1

Kudos [?]: 2 [0], given: 0

Padma, I did the traditional way, lets say 1,2,3,4,5,Ben and Ann are 7 contestants. First Ben and Ann are ruled out, we r left with 1,2,3,4,5 and so the possible selections are 1234, 1235, 1245,1345 and 2345.
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

Well this is tricky. It lets you comeout with 7C4- 2 * 5C2 = 15
But what we need is 4 contestants out of 5 since we are to exclude A&B
The answer is simply 5C4 = 5

I wish stoolfi has some innovative method to explain the discrepency here.
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

Well I have the innovative way. I hate to depart from classical way to solve this.

Let us say we have chosen A & B now now we choose 2 out of remaining 5
we get a pair and let us call it C
so we have A,B,C we can arrange these in 3 ways
(A,B,C), (C,A,B), (B,C,A)
so we have 3 * 5C2 combinations in which A and B are present = 30
Total combinations = 7C4 = 35

Desired combinations = 35-30 =5
Intern
Joined: 05 Jan 2004
Posts: 27
Location: Los Angeles
Followers: 0

Kudos [?]: 0 [0], given: 0

We do not want Ben and Ann, which are 2 contestants, they could be any 2 that we do not want

So we are left with only 5 others

So we got to choose 4 out of them

5C4 = 5!/4! = 5 ways that you can choose
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