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B
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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
A. 14
B. 18
C. 35
D. 56
E. 144
13 Jul 2009, 14:00
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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:
\(= 6 \times 3\)
\(= 18\)
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:
\(= 6 \times 3\)
\(= 18\)
10 Dec 2013, 06:07
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aeglorre
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.
Theory on Combinations: math-combinatorics-87345.html
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Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.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
