Number of ways to seat 6 boys and 6 girls in a row so that no two girls sit together

Introduction / Context:
This question tests advanced permutation reasoning with spacing restrictions. You must seat 6 boys and 6 girls in a single row such that no two girls sit next to each other. Problems like this are common in competitive exams and require a clear strategy.

Given Data / Assumptions:

• There are 6 distinct boys.
• There are 6 distinct girls.
• All 12 children must be seated in one row.
• No two girls are allowed to occupy adjacent seats.
• All boys and girls are distinguishable from one another.

Concept / Approach:
The standard approach is to first seat the boys, then place the girls in the gaps between or at the ends of boys. Once the boys are placed, they create fixed slots where girls can be inserted. We then choose positions for girls in these slots and arrange the girls.

Step-by-Step Solution:
Step 1: Arrange the 6 boys in a row in 6! ways.Step 2: After seating boys, there are 7 possible slots where girls can sit: one before the first boy, one after the last boy, and one between every pair of consecutive boys.Step 3: To ensure that no two girls sit together, we can place at most one girl in each slot.Step 4: We have 6 girls and 7 distinct slots, so we must choose 6 different slots from these 7 and then assign the girls to those chosen slots.Step 5: Number of ways to pick ordered slots for the 6 distinct girls is 7P6, that is, permutations of 7 slots taken 6 at a time.Step 6: Therefore, the total number of valid seating arrangements is 6! * 7P6.

Verification / Alternative check:
You can also compute 7P6 as 7! / 1! = 5040 and note that 6! = 720, so the total number of arrangements equals 720 * 5040 which is 3628800. This is a very large but reasonable number given 12 distinct people with restrictions.

Why Other Options Are Wrong:

• 2(6!) and 6! x 7 do not incorporate the full complexity of distributing 6 distinct girls into 7 slots.
• None is incorrect because the expression 6! x 7P6 correctly models the situation.

Common Pitfalls:
Students sometimes mistakenly choose only 6 slots out of 7 using combinations 7C6 and then multiply by 6!, which in fact already equals 7P6 but is sometimes misinterpreted. Others forget the spacing idea and attempt to directly arrange all 12 people with complicated inclusion exclusion. The slot method is far simpler and more reliable.

Final Answer:
The number of ways to seat the group so that no two girls sit together is 6! times 7P6, so the correct answer is 6! x ⁷P₆.