(380 - 9)/(4,000 - 54) x (390 - 9)/(4,000 - 55) x (400 - 9)/(4,000 - 56) x (390 - 9)/(4,000 - 57) x (1,210 - 9)/(4,000 - 58) x (1,230 - 9)/(4,000 - 59) x 6!

Its conditional probability though

Let A = first 54 students consist of 9 students from each of the 6 different levels

Let B = next 6 students from the remaining student population are each from the 6 different levels.

P(B|A) = P(B n A) / P(A)
P(B|A) = [P(B) x P(A)] / P(A)
P(B|A) = P(B)