Question

5 boys and 6 girls are arranged in all possible ways. Let \(X\) denote the number of linear arrangements in which no two boys sit together, and \(Y\) denote the number of linear arrangements in which no two girls sit together. If \(Z\) denotes the number of ways of arranging all of them around a circular table such that no two boys sit together, then \(X:Y:Z\) = ?

• A. \(1:1:21\)
• B. \(21:1:1\)
• C. \(7:5:5\)
• D. \(4:3:3\)

Hint:

In problems involving seating arrangements where no two specific groups can sit together, always arrange one group first and then place the other group in the available gaps.

Answer 1

The Correct Option is B

Step 1: Finding \(X\) (Linear arrangement where no two boys sit together) To ensure that no two boys sit together in a linear arrangement, we first arrange the 6 girls: \[ 6! = 720 \] Now, we place the 5 boys in the 7 available gaps: \[ \text{Ways to arrange 5 boys in 7 gaps} = \binom{7}{5} = \frac{7!}{5!(7-5)!} = 21 \] Arranging 5 boys: \[ 5! = 120 \] Thus, \[ X = 6! \times \binom{7}{5} \times 5! = 720 \times 21 \times 120 \]
Step 2: Finding \(Y\) (Linear arrangement where no two girls sit together) By similar reasoning, the number of ways to arrange them when no two girls sit together: \[ Y = 5! \times \binom{7}{6} \times 6! = 5! \times 7 \times 6! = 120 \times 7 \times 720 \] Since \( X = 21 Y \), we get the ratio \( X:Y = 21:1 \).
Step 3: Finding \(Z\) (Circular arrangement where no two boys sit together) In a circular arrangement, we fix one girl as a reference point, arranging the remaining 5 girls: \[ 5! = 120 \] The 5 boys are placed in the 6 gaps: \[ 5! = 120 \] Thus, \[ Z = 5! \times 5! = 120 \times 120 \] Since \( X = 21Z \) and \( Y = Z \), we get: \[ X:Y:Z = 21:1:1 \] Thus, the final answer is: \[ \boxed{21:1:1} \]