If we increase something by 200%, we are multiplying it by 3.
There is a conceptual solution here, but it's more complicated than backsolving, so let's just test answers. If you have, say, 10 students, 5 boys and 5 girls, there will be two ways to arrange them:
BGBGBGBGBG
or
GBGBGBGBGB
In the first case, you'd have 5 choices for the first boy, 5 for the first girl, 4 for the second boy, 4 for the second girl, and so on. So you'd have 5! * 5! arrangements in total. The same is true in the second case, so we can arrange the students in 5!*5! + 5!*5! = 2*5!*5! ways. From here you can see how to count arrangements for any even number of students - if we had, say, 12 students, the answer would be 2*6!*6!
When we have an odd number of students, so say we have 6 boys and 5 girls, then we must put boys at each end of the arrangement, so it must look like:
BGBGBGBGBG
and we find we have 6!*5! arrangements in total. From here you can see how to count whenever n is odd; with 13 students, say, the answer would be 7! * 6!
So it's possible to work out fairly quickly the number of arrangements for any value of n. Notice that when n is 10, we have 2*5!*5! arrangements, and when n=11, we have 6!*5! = 6*5!*5! arrangements. Since 6*5!*5! is exactly 3 times 2*5!*5!, the answer is 10.
The question is too long to be a realistic GMAT problem though.
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