Intro to statistics question (probability) A fleet of nine taxis is dispatched to three airports, in such a way that three go to airport A, five go to airport B and one goes to airport C.
If exactly three taxis are in need of repair. What is the probability that every airport receives one of the taxis requiring repairs.
My method was total number of ways is (9!)/(3!*5!*1!)= 504 (this is correct)
Now you have three groups and of every group one position is already filled. So: ....R, ..R, R
So the numbers of ways to distribute the rest is: (6!)/(2!*4!)=15
But according to answer model of my professor, it should be 90. So I assume he says the number of ways of distrubiting these 3 broken cabs over the 3 spots is 3*2*1.
But to me this seems wrong?
The broken taxis don't differ so R(1)R(2)R(3) is same as R(2)R(1)R(3).
If you standing at the airport A, it doesn't matter which of the broken cabs stands there?
 A: A probability is usually between $0$ and $1$.  
Try a simpler question: two airports, namely D with two taxis and E with two.   Suppose the taxis are $T_1$, $T_2$, $T_3$, $T_4$ and the even numbers are broken.
The four equally probably cases with a broken taxi at each airport are 


*

*$\{T_1,T_2\}, \{T_3,T_4\}$

*$\{T_1,T_4\}, \{T_2,T_3\}$

*$\{T_2,T_3\}, \{T_1,T_4\}$

*$\{T_3,T_4\}, \{T_1,T_2\}$


The two equally probably cases with both broken taxis at the same airport are


*

*$\{T_1,T_3\}, \{T_2,T_4\}$

*$\{T_2,T_4\}, \{T_1,T_3\}$


Your method would give $\dfrac{4!}{2!\,2!}=6$ ways of distributing the four taxis and $\dfrac{2!}{1!\,1!}=2$ ways with one broken taxi at each airport, making the probability $\dfrac26=\dfrac13$.  Too small. 
If instead we multiplied by $2!$ because there are two broken taxis, that would make the probability $\dfrac46=\dfrac23$.  Just right.
A: Comment: Please see my first comment above. 
Below is a simulation in R statistical software of a million performances
of the experiment, in which 1's denote bad taxis and 0's good ones. The
simulated answer should be correct to two or three places.
m = 10^6;  txi = c(1,1,1,0,0,0,0,0,0)
a1 = a2 = a3 = numeric(m)
for (i in 1:m) {
  prm = sample(txi, 9)
  a1[i] = sum(prm[1:3])  # nr bad taxis at Airport 1
  a2[i] = sum(prm[4:8])  # etc
  a3[i] = sum(prm[9]) }
mean(a1==1 & a2==1 & a3==1)
## 0.179161
90/504
## 0.1785714

It seems your professor is correct this time. You should think
carefully about what kind of outcomes are included among the 504.
Peerhaps @Henry's Answer (+1) to a similar, simpler problem has given you a clue how to do that.
