Step 1: Understanding the Concept:
This question involves comparing two different scenarios of selecting students. Both scenarios use the principles of combinations and the fundamental counting principle. We need to calculate the number of ways for each column and then compare them.
Step 2: Key Formula or Approach:
The number of ways to choose \(r\) items from a set of \(n\) distinct items is given by the combination formula:
\[ C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
The fundamental counting principle states that if one event can occur in \(m\) ways and a second event can occur in \(n\) ways, then the two events can occur in \(m \times n\) ways.
Step 3: Detailed Explanation:
For Column A:
We need to choose 2 students from a class of 36. The order of selection does not matter, so we use combinations.
Number of ways = \( C(36, 2) \)
\[ C(36, 2) = \frac{36!}{2!(36-2)!} = \frac{36!}{2!34!} = \frac{36 \times 35}{2 \times 1} = 18 \times 35 = 630 \]
So, there are 630 ways to choose two monitors from the class.
For Column B:
We need to choose one monitor from 18 girls and one monitor from 18 boys.
Number of ways to choose 1 girl from 18 girls = \( C(18, 1) = 18 \).
Number of ways to choose 1 boy from 18 boys = \( C(18, 1) = 18 \).
Using the fundamental counting principle, the total number of ways to choose one girl AND one boy is:
\[ \text{Total ways} = 18 \times 18 = 324 \]
So, there are 324 ways to choose one girl and one boy as monitors.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 630
Quantity B = 324
Clearly, Quantity A is greater than Quantity B.