Four boys and three girls line up for an interview. What is the probability that they stand in strictly alternating gender order (B–G–B–G–B–G–B)?

Introduction / Context:
Alternation constraints are classic permutation-with-pattern problems. With 4 boys and 3 girls, a perfect alternation is only possible in the pattern B–G–B–G–B–G–B (starting and ending with boys).

Given Data / Assumptions:

• Distinct individuals: 4 boys, 3 girls.
• Total permutations without restriction = 7!.
• Alternation pattern forced: B–G–B–G–B–G–B.

Concept / Approach:
Place boys in the 4 boy slots in 4! ways, and girls in the 3 girl slots in 3! ways. Divide by 7! to get the probability.

Step-by-Step Solution:
Favourable arrangements = 4! * 3! = 24 * 6 = 144.Total arrangements = 7! = 5040.Probability = 144 / 5040 = 1 / 35.

Verification / Alternative check:
Since the counts differ by 1, only a single alternation pattern is feasible; the computed probability matches that unique pattern count.

Why Other Options Are Wrong:
1/34 and 1/17 are not reducible forms from factorial counts; 1/68 would correspond to an extra symmetry that does not exist.

Common Pitfalls:
Allowing a starting girl pattern (impossible with 4 boys and 3 girls) or double-counting placements.

Final Answer:
1/35