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:

The value of \(C(n, r)\) grows faster when \(n\) is larger. In this problem, Quantity A involves selecting from larger pools for the larger required group (\(C(25,5)\)) compared to Quantity B (\(C(10,5)\)). This suggests that Quantity A will be larger.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a combination problem where a group is formed by selecting members from two distinct pools (boys and girls). The total number of ways to form the party is the product of the number of ways to select the boys and the number of ways to select the girls.
Step 2: Key Formula or Approach:
The number of ways to choose \(r\) items from a set of \(n\) is \(C(n, r) = \frac{n!}{r!(n-r)!}\). Total ways = (Ways to choose boys) \( \times \) (Ways to choose girls).
Step 3: Detailed Explanation:
For Column A:
We need to select 5 boys from 25 and 3 girls from 10. - Ways to choose 5 boys from 25 = \(C(25, 5)\). - Ways to choose 3 girls from 10 = \(C(10, 3)\). Total ways for party A = \(C(25, 5) \times C(10, 3)\).
For Column B:
We need to select 5 boys from 10 and 3 girls from 25. - Ways to choose 5 boys from 10 = \(C(10, 5)\). - Ways to choose 3 girls from 25 = \(C(25, 3)\). Total ways for party B = \(C(10, 5) \times C(25, 3)\).
Comparison:
We are comparing A = \(C(25, 5) \times C(10, 3)\) with B = \(C(10, 5) \times C(25, 3)\). Let's analyze the ratio A/B: \[ \frac{A}{B} = \frac{C(25, 5) \times C(10, 3)}{C(10, 5) \times C(25, 3)} \] Let's expand the terms: \[ C(25, 5) = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1} \quad C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \] \[ C(10, 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \quad C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \] Let's use the factorial form for easier cancellation. \[ \frac{A}{B} = \frac{\frac{25!}{5!20!} \times \frac{10!}{3!7!}}{\frac{10!}{5!5!} \times \frac{25!}{3!22!}} = \frac{25! \cdot 10! \cdot 5! \cdot 5! \cdot 3! \cdot 22!}{5! \cdot 20! \cdot 3! \cdot 7! \cdot 10! \cdot 25!} = \frac{5! \cdot 22!}{20! \cdot 7!} \] \[ \frac{A}{B} = \frac{5! \times (22 \times 21 \times 20!)}{20! \times (7 \times 6 \times 5!)} = \frac{22 \times 21}{7 \times 6} = \frac{462}{42} = 11 \] Since \(\frac{A}{B} = 11\), which is greater than 1, it means A is 11 times larger than B.
Step 4: Final Answer:
Quantity A is greater than Quantity B.