Question

• A. The quantity on the left is greater
• B. The quantity on the right is greater
• C. Both are equal
• D. The relationship cannot be determined without further information

Hint:

Instead of full calculation, you can compare the terms. A = \(27 \times C(14, 2) = 27 \times \frac{14 \times 13}{2} = 27 \times 7 \times 13\). B = \(C(27, 2) \times 14 = \frac{27 \times 26}{2} \times 14 = 27 \times 13 \times 14\). Comparing A and B, we are comparing \(7\) and \(14\). Since \(14>7\), B is greater.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a combination problem involving two independent selections: choosing boys from a group of boys and choosing girls from a group of girls. The total number of ways is the product of the number of ways for each independent selection.
Step 2: Key Formula or Approach:
The number of ways to choose \(r\) items from a set of \(n\) is given by the combination formula: \(C(n, r) = \frac{n!}{r!(n-r)!}\). We use the multiplication principle to find the total number of ways.
Step 3: Detailed Explanation:
For Column A:
Select 1 boy from 27 boys AND 2 girls from 14 girls. Ways to select 1 boy from 27: \[ C(27, 1) = \frac{27!}{1!(27-1)!} = 27 \] Ways to select 2 girls from 14: \[ C(14, 2) = \frac{14!}{2!(14-2)!} = \frac{14 \times 13}{2 \times 1} = 7 \times 13 = 91 \] Total ways = (Ways to select boys) \( \times \) (Ways to select girls) \[ \text{Total ways} = 27 \times 91 = 2457 \] So, Quantity A is 2457.
For Column B:
Select 2 boys from 27 boys AND 1 girl from 14 girls. Ways to select 2 boys from 27: \[ C(27, 2) = \frac{27!}{2!(27-2)!} = \frac{27 \times 26}{2 \times 1} = 27 \times 13 = 351 \] Ways to select 1 girl from 14: \[ C(14, 1) = \frac{14!}{1!(14-1)!} = 14 \] Total ways = (Ways to select boys) \( \times \) (Ways to select girls) \[ \text{Total ways} = 351 \times 14 = 4914 \] So, Quantity B is 4914.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 2457
Quantity B = 4914
Quantity B is greater than Quantity A.