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

 It is currently 30 Aug 2016, 16:19

### 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

# A 4-person task force is to be formed from the 4 men and 3 w

Author Message
TAGS:

### Hide Tags

Intern
Joined: 10 Jul 2009
Posts: 43
Location: Beijing
Followers: 1

Kudos [?]: 16 [1] , given: 3

A 4-person task force is to be formed from the 4 men and 3 w [#permalink]

### Show Tags

13 Jul 2009, 12:19
1
KUDOS
1
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

90% (01:37) correct 10% (01:09) wrong based on 198 sessions

### HideShow timer Statistics

A 4-person task force is to be formed from the 4 men and 3 women who work in Company G's human resources department. If there are to be 2 men and 2 women on this task force, how many different task forces can be formed?

A. 14
B. 18
C. 35
D. 56
E. 144
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Dec 2013, 06:00, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
Current Student
Joined: 03 Aug 2006
Posts: 116
Followers: 4

Kudos [?]: 219 [5] , given: 3

### Show Tags

13 Jul 2009, 14:00
5
KUDOS
1
This post was
BOOKMARKED

This is a combinations problem.

$$C^n_k = \frac{n!}{k!(n-k)!}$$

where $$n$$ = total number of items available to select and $$k$$ = number of items to be selected.

In this problem there are two distinct sets: 1 set of men and 1 set of women.

The number of ways we can select 2 ($$k$$) men from a total of 4 ($$n$$) men:

$$\Rightarrow C^n_k = \frac{n!}{k!(n-k)!}$$

$$\Rightarrow C^4_2 = \frac{4!}{2!(4-2)!}$$

$$\Rightarrow C^4_2 = \frac{4!}{2!2!}$$

$$\Rightarrow C^4_2 = \frac{4 \times 3}{2 \times 1} = 6$$

Similarly the number of ways we can select 2 ($$k$$) women from a total of 3 ($$n$$) women:

$$\Rightarrow C^n_k = \frac{n!}{k!(n-k)!}$$

$$\Rightarrow C^3_2 = \frac{3!}{2!(3-2)!}$$

$$\Rightarrow C^3_2 = \frac{3!}{2!1!}$$

$$\Rightarrow C^3_2 = 3$$

Total number of different task forces of 2 men and 2 women possible:

$$= 6 \times 3$$

$$= 18$$
Senior Manager
Joined: 25 Mar 2009
Posts: 305
Followers: 7

Kudos [?]: 184 [1] , given: 6

### Show Tags

13 Jul 2009, 14:37
1
KUDOS
nookway's calcs are good.

Just one note about why we are using the nCr formula.

Let's label the 4 men M1, M2, M3, and M4. One possible combo is if you first choose M1, then choose M2. This team of M1 and M2 is the same as if you had chosen M2 first, then M1 second. So since M1M2 is not distinct from M2M1, order does not matter. When order does not matter, use the combination formula.
Intern
Joined: 10 Jul 2009
Posts: 43
Location: Beijing
Followers: 1

Kudos [?]: 16 [0], given: 3

### Show Tags

13 Jul 2009, 14:39
Thanks a lot!

I was looking at it as being much more complicated than it actually was.

Sorry, the OA was indeed 18.
Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 48 [0], given: 16

### Show Tags

03 Jul 2011, 01:20
do we need to think as they are two independent events and
to select 2 seats from 4 men is 6 and similarly to filll 2 seats from 3 women is 3....
so it is 6X3 = 18?
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Manager
Joined: 12 Jan 2013
Posts: 244
Followers: 4

Kudos [?]: 58 [0], given: 47

### Show Tags

10 Dec 2013, 05:51
Hi,

wouldn't it suffice if we simply used 4!/2! for men and added that to 3!/1! for women?

4!/2! = 4x3 = 12
3!/1! = 6

12 + 6 = ways 4 men can fill 2 positions + ways 3 women can fill 2 positions = 18.

Why would this NOT work in general (because it works in this specific case), and why would we need to use relatively "complicated" divisions nCr formulas when the solution is much more straightforward than that?
Math Expert
Joined: 02 Sep 2009
Posts: 34511
Followers: 6303

Kudos [?]: 79986 [1] , given: 10022

### Show Tags

10 Dec 2013, 06:07
1
KUDOS
Expert's post
aeglorre wrote:
A 4-person task force is to be formed from the 4 men and 3 women who work in Company G's human resources department. If there are to be 2 men and 2 women on this task force, how many different task forces can be formed?

A. 14
B. 18
C. 35
D. 56
E. 144

Hi,

wouldn't it suffice if we simply used 4!/2! for men and added that to 3!/1! for women?

4!/2! = 4x3 = 12
3!/1! = 6

12 + 6 = ways 4 men can fill 2 positions + ways 3 women can fill 2 positions = 18.

Why would this NOT work in general (because it works in this specific case), and why would we need to use relatively "complicated" divisions nCr formulas when the solution is much more straightforward than that?

What is the logic behind 4!/2! and 3!/1!?

The number of ways to choose 2 men out of 4 is $$C^2_4=6$$;
The number of ways to choose 2 women out of 3 is $$C^2_3=3$$.

Principle of Multiplication says that if one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways.

Thus the number of ways to choose 2 men AND 2 women is 6*3=18.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________
Intern
Joined: 19 Mar 2013
Posts: 23
Followers: 0

Kudos [?]: 3 [0], given: 27

Re: A 4-person task force is to be formed from the 4 men and 3 w [#permalink]

### Show Tags

12 Dec 2013, 06:46
we can chose 2 women out of 3 in 3C2 ways = 3
2 men out of 4 in 4C2 ways = 6

different combinations of 6 men and 3 women = 6*3= 18
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11158
Followers: 512

Kudos [?]: 134 [0], given: 0

Re: A 4-person task force is to be formed from the 4 men and 3 w [#permalink]

### Show Tags

14 Mar 2015, 20:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: A 4-person task force is to be formed from the 4 men and 3 w   [#permalink] 14 Mar 2015, 20:05
Similar topics Replies Last post
Similar
Topics:
2 From a total of 5 boys and 4 girls, how many 4-person committees can 5 26 Mar 2015, 04:21
7 Find the probability that a 4 person committee chosen at ran 5 14 Dec 2009, 21:05
10 In a 4 person race, medals are awarded to the fastest 3 runn 13 03 Dec 2009, 07:52
3 In a 4 person race, medals are awarded to the fastest 3 runn 6 20 Nov 2009, 02:53
43 In a 4 person race, medals are awarded to the fastest 3 runn 10 18 Sep 2007, 23:32
Display posts from previous: Sort by