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

The answer is B.

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:

\(= 6 \times 3\)

\(= 18\)

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.

Answer: B.


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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.

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

Answer: B
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Thanks a lot!

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

Sorry, the OA was indeed 18.

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?

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?

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

# of taskforces = \(4C2*3C2 = 6*3 = 18\)

IMO B

Number of ways to choose 2 men out of 4: 4C2 = 6
Number of ways to choose 2 women out of 3: 3C2 = 3

6 * 3 = 18 different task forces

Answer is B.
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