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

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Intern
Joined: 10 Jul 2009
Posts: 41
Location: Beijing
A 4-person task force is to be formed from the 4 men and 3 w  [#permalink]

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Updated on: 10 Dec 2013, 06:00
5
5
00:00

Difficulty:

5% (low)

Question Stats:

89% (01:01) correct 11% (01:43) wrong based on 326 sessions

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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

Originally posted by optiquezt on 13 Jul 2009, 12:19.
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.
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Joined: 03 Aug 2006
Posts: 80

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13 Jul 2009, 14:00
8
5

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:

$$= 6 \times 3$$

$$= 18$$
##### General Discussion
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Joined: 25 Mar 2009
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13 Jul 2009, 14:37
1
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.
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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.
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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?
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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?
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10 Dec 2013, 06:07
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3
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

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

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

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06 Jul 2019, 03:27
How to solve this question using slot method? Bunuel

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

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11 Jul 2019, 20:04
optiquezt 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

We need to select 2 men from a group of 4 men. Since order does not matter, we can use combinations.

4C2 = (4 x 3)/2! = 12/2 = 6

Similarly, the number of ways to select 2 women from 3 is:

3C2 = (3 x 2)/2! = 6/2 = 3

Finally, the number of ways we can choose 2 men from 4 and 2 women from 3 is:

4C2 x 3C2 = 6 x 3 = 18

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

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11 Jul 2019, 22:04
optiquezt 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

# of taskforces = $$4C2*3C2 = 6*3 = 18$$

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

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08 Oct 2019, 08:00
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Re: A 4-person task force is to be formed from the 4 men and 3   [#permalink] 08 Oct 2019, 08:00
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