Of the 12 temporary employees in a certain company, 4 will : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 12:12

### 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

# Of the 12 temporary employees in a certain company, 4 will

Author Message
Manager
Joined: 17 Jan 2006
Posts: 92
Followers: 1

Kudos [?]: 17 [0], given: 0

Of the 12 temporary employees in a certain company, 4 will [#permalink]

### Show Tags

30 May 2006, 16:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary are women. how many possible groups of 4 temporary employees consist of 3 women an 1 man

22
35
56
70
105
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 10

Kudos [?]: 60 [0], given: 0

### Show Tags

30 May 2006, 16:58
dinesh8 wrote:
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary are women. how many possible groups of 4 temporary employees consist of 3 women an 1 man?

22
35
56
70
105

=5c3x7c1
= 10x7
=70
Senior Manager
Joined: 08 Jun 2004
Posts: 497
Location: Europe
Followers: 1

Kudos [?]: 72 [0], given: 0

### Show Tags

30 May 2006, 18:55
Agree with Prof, 70 it is.
Manager
Joined: 12 Feb 2006
Posts: 115
Followers: 1

Kudos [?]: 68 [0], given: 0

### Show Tags

30 May 2006, 20:41
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!
SVP
Joined: 30 Mar 2006
Posts: 1737
Followers: 1

Kudos [?]: 77 [0], given: 0

### Show Tags

30 May 2006, 21:06
Ans - 70

5C3*7C1 = 10*7 = 70
Director
Joined: 09 Oct 2005
Posts: 720
Followers: 3

Kudos [?]: 23 [0], given: 0

### Show Tags

30 May 2006, 21:08
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120
_________________

IE IMBA 2010

SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5

Kudos [?]: 74 [0], given: 0

### Show Tags

30 May 2006, 21:21
Yurik79 wrote:
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120

5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10
6C4=(6*5*4!)/(4!2!) = 15
10C3=(10*9*8x7!)/(7!x3x2x1) = 120
Manager
Joined: 12 Feb 2006
Posts: 115
Followers: 1

Kudos [?]: 68 [0], given: 0

### Show Tags

31 May 2006, 19:57
Thanks Everyone,

Just so I'm 100% clear, what is your approach if there are 7 women and 5 men in this question?

Thanks,
Director
Joined: 09 Oct 2005
Posts: 720
Followers: 3

Kudos [?]: 23 [0], given: 0

### Show Tags

31 May 2006, 21:25
HIMALAYA wrote:
Yurik79 wrote:
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120

5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10
6C4=(6*5*4!)/(4!2!) = 15
10C3=(10*9*8x7!)/(7!x3x2x1) = 120

Yes I know the formula ))Just simplified a little
in 6C4=(6*5*4!)/(4!2!) = 15 4! is cancelled anyway
5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10 same here 3! is canceled
_________________

IE IMBA 2010

Manager
Joined: 11 Oct 2005
Posts: 94
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

06 Jun 2006, 12:02
A little late - 5C3*7C1 = 70
06 Jun 2006, 12:02
Display posts from previous: Sort by