Okay, so I need to figure out how many ways 6 girls and 2 boys can sit in a row with the condition that the two boys must sit next to each other. Let me start by understanding the problem.

First, if there were no restrictions, the total number of ways to arrange 8 people (6 girls + 2 boys) would be 8 factorial, which is 8! = 40320. But here, there's a restriction: the two boys must be sitting next to each other. So I need to adjust for that.

When two people must sit together, a common strategy is to treat them as a single unit or "block." So instead of having 8 separate individuals, we can think of the two boys as one unit plus the 6 girls, making a total of 7 units. The number of ways to arrange these 7 units would be 7! (7 factorial). But wait, within this block, the two boys can switch places. So for each arrangement, there are 2! ways to arrange the boys within their block. Therefore, the total number of arrangements should be 7! * 2!.

Let me verify this. If we consider the two boys as a single entity, we have 7 entities to arrange. Each of these entities can be arranged in 7! ways. Since the two boys can be in two different orders within their block, we multiply by 2!. So yes, 7! * 2! = 5040 * 2 = 10080.

But let me check if there's another way to approach this problem to confirm. Maybe using the concept of treating the two boys as adjacent and calculating the number of possible positions for them.

The total number of seats is 8. If the two boys need to sit next to each other, they can occupy positions 1&2, 2&3, 3&4, 4&5, 5&6, 6&7, or 7&8. So there are 7 possible adjacent pairs. For each of these pairs, the two boys can be arranged in 2! ways (AB or BA). Then, the remaining 6 girls can be arranged in the remaining 6 seats in 6! ways. So the total number of arrangements would be 7 (positions) * 2! (arrangements of boys) * 6! (arrangements of girls) = 7 * 2 * 720 = 7 * 1440 = 10080. That's the same result as before. So both methods give the same answer, which is reassuring.

Another way to think about it is using permutations with restrictions. The formula when two specific people must be adjacent is (n-1)! * 2!, where n is the total number of people. Wait, in this case, n=8, so (8-1)! * 2! = 7! * 2! = 5040 * 2 = 10080. Yep, same answer again.

Alternatively, using the inclusion-exclusion principle. The total number of arrangements without restrictions is 8!. Then subtract the number of arrangements where the two boys are not sitting next to each other. But calculating the complement might be more complicated here because you have to subtract all the cases where the boys are separated. Whereas the original method of treating them as a block is straightforward.

So all three methods converge on 10080. Therefore, I'm confident that the answer is 10080.

But let me just recap to make sure I didn't skip any steps. When dealing with two people that must be together, consider them as a single entity. Then, calculate the permutations of all entities, then multiply by the permutations within the entity. That's exactly what I did. So 7! for the 7 entities (6 girls + 1 boy pair) and 2! for the two boys. Multiply them together. Yep, that's solid.

Another sanity check: if there were only 2 people, the number of ways they can sit next to each other is 2! = 2, which makes sense. If there are 3 people and two must sit together, treating them as a block gives 2! * 2! = 4. Let's see: the possible arrangements are AB C, BA C, C AB, C BA. So four arrangements, which is correct. So scaling up, in the original problem, the formula works.

Therefore, I think the answer is definitely 10080.
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