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:

When a variable is missing from a quantitative comparison question, check if the relationship between the two columns changes depending on the possible values of that variable. If it does, the answer is (D).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem asks for the number of ways to form a committee by selecting from two groups (professors and students). However, a critical piece of information, the total number of students to choose from, is missing. We need to analyze if a comparison is possible without this information.
Step 2: Key Formula or Approach:
The total ways to form the committee is (Ways to choose professors) \( \times \) (Ways to choose students). We will use the combination formula \(C(n, r)\). Let 'S' be the total number of students.
Step 3: Detailed Explanation:
In both columns, the number of ways to choose 2 professors from 10 is the same: \[ C(10, 2) = \frac{10 \times 9}{2} = 45 \] The comparison depends entirely on the number of ways to choose the students. Let 'S' be the total number of shortlisted students. For the combinations to be possible, \(S\) must be at least 4 (since in Column A, one is excluded and we still need to choose 3).
For Column A: A particular student is excluded
- We need to choose 3 students. - The pool of students is reduced by one, so we choose from \(S-1\) students. - Ways to choose students = \(C(S-1, 3)\). - Total ways for A = \(45 \times C(S-1, 3)\).
For Column B: A particular student is included
- We need to choose 3 students, and one is already chosen. - We need to choose \(3-1 = 2\) more students. - The pool of students is reduced by one (the one already included), so we choose from \(S-1\) students. - Ways to choose students = \(C(S-1, 2)\). - Total ways for B = \(45 \times C(S-1, 2)\).
Comparison:
We are comparing \(C(S-1, 3)\) with \(C(S-1, 2)\). \[ C(S-1, 3) = \frac{(S-1)!}{3!(S-4)!} \quad \text{and} \quad C(S-1, 2) = \frac{(S-1)!}{2!(S-3)!} \] The relationship between them is \(C(n, k) = \frac{n-k+1}{k} C(n, k-1)\). So, \(C(S-1, 3) = \frac{(S-1)-3+1}{3} C(S-1, 2) = \frac{S-3}{3} C(S-1, 2)\). - If \(S-3>3\) (i.e., \(S>6\)), then \(C(S-1, 3)>C(S-1, 2)\) and Quantity A is greater. - If \(S-3<3\) (i.e., \(S<6\)), then \(C(S-1, 3)<C(S-1, 2)\) and Quantity B is greater. (e.g., if S=5, A=45*C(4,3)=180, B=45*C(4,2)=270). - If \(S-3 = 3\) (i.e., \(S = 6\)), then \(C(S-1, 3) = C(S-1, 2)\) and the quantities are equal. (A=45*C(5,3)=450, B=45*C(5,2)=450). Since the value of S is not given, we cannot determine which quantity is greater.
Step 4: Final Answer:
The relationship cannot be determined from the information given.