Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Aug 2016, 21:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Ben and Ann are among 7 contestants from which 4

Author Message
Director
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 75 [0], given: 0

Ben and Ann are among 7 contestants from which 4 [#permalink]

### Show Tags

28 Jan 2004, 16:58
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

### Show Tags

28 Jan 2004, 17:57
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: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 83 [0], given: 0

### Show Tags

29 Jan 2004, 11:52
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: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 83 [0], given: 0

### Show Tags

29 Jan 2004, 12:00
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

### Show Tags

29 Jan 2004, 12:10
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
Display posts from previous: Sort by