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:

For "never together" problems, the gap method is the most reliable approach. Arrange the unrestricted group first, then place the restricted group in the gaps. The number of gaps is always one more than the number of items in the unrestricted group.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a classic permutation problem with a restriction: "no two girls may sit together". We solve this using the "gap method". First, we arrange the items without the restriction (boys) and then place the restricted items (girls) in the gaps created. The phrasing in Column A, "6 boys and girls", is ambiguous. A common interpretation in competitive exams for such phrasing is "6 boys and 6 girls". We will proceed with this assumption.
Step 2: Key Formula or Approach:
1. Arrange the \(n\) boys in \(n!\) ways. This creates \(n+1\) gaps.
2. Choose \(r\) gaps for the \(r\) girls and arrange them in \(P(n+1, r)\) ways.
3. Total ways = (Ways to arrange boys) × (Ways to place girls) = \(n! \times P(n+1, r)\).
Step 3: Detailed Explanation:
For Column A (Assuming 6 boys and 6 girls):
- Step 1: Arrange the 6 boys.
The number of ways to arrange 6 boys in a row is \(6!\).
6! = 720
- Step 2: Place the 6 girls in the gaps.
Arranging the 6 boys creates 7 gaps (_B_B_B_B_B_B_). We need to place the 6 girls in these 7 gaps. The number of ways to do this is \(P(7, 6)\).
P(7, 6) = \( \frac{7!}{(7-6)!} \) = 7! = 5040
- Step 3: Calculate the total ways.
Total ways = 6! × P(7, 6) = 720 × 5040 = 3,628,800
So, Quantity A is 3,628,800.
For Column B (5 boys and 3 girls):
- Step 1: Arrange the 5 boys.
The number of ways to arrange 5 boys in a row is \(5!\).
5! = 120
- Step 2: Place the 3 girls in the gaps.
Arranging the 5 boys creates 6 gaps (_B_B_B_B_B_). We need to place the 3 girls in these 6 gaps. The number of ways to do this is \(P(6, 3)\).
P(6, 3) = \( \frac{6!}{(6-3)!} \) = 6 × 5 × 4 = 120
- Step 3: Calculate the total ways.
Total ways = 5! × P(6, 3) = 120 × 120 = 14,400
So, Quantity B is 14,400.
Step 4: Final Answer:
Comparing the two quantities:
Quantity A = 3,628,800
Quantity B = 14,400
Quantity A is greater than Quantity B.