Ten students are waiting to board a school bus. "Student 1"

I think you are making an assumption that the three students need to get in immediately next to each other, other students can get into the bus in between S1-S3, S3-S2 or S1-S2.

Overall, student 2 can take any place starting from 3 to 10
Given below are S2 positions -> corresponding combinations to seat S1 & S3 (Multiplied by 2 is used because their positions are interchangeable)
S2=3 -> S1, S3 = 2C2 * 2 = 2
S2=4 -> S1, S3 = 3C2 * 2 = 6
S2=5 -> S1, S3 = 4C2 * 2 = 12
S2=6 -> S1, S3 = 5C2 * 2 = 20
S2=7 -> S1, S3 = 6C2 * 2 = 30
S2=8 -> S1, S3 = 7C2 * 2 = 42
S2=9 -> S1, S3 = 8C2 * 2 = 56
S2=10 -> S1, S3 = 9C2 * 2 = 72

Total positions for S1, S3 = 240
in each of these 240 positions, other 7 kids can be seated in 7! ways
240*7! = 12,09,600

That's option C

Of course, doesn't look like one can solve this within 2 minutes.

Let me know if you have a different approach.

This is a wonderful explanation.. deserve a kudos

Posted from my mobile device

Three (3) places can be chosen out of 10 places in 10C3 ways.

There are only two (2) possible arrangements: S1S3S2 or S3S1S2

The remaining seven (7) places can be arranged in 7! ways.

Therefore, the total number of ways in which boarding can occur equals the product:

10C3 X 2 X 7! = 1209600

ANSWER: (C)