"A fleet of ten taxis is to be dispatched to three locations in such a way that three taxis go to the first location, five taxis go to the second location, and two taxis go to the third location. (a) In how many distinct ways can the taxis be dispatched?"
I know in the first part to use n!/(n1!n2!n3!). So in this problem it's 10!/(3!5!2!)=2520
It's part (b) that I can't figure out: "Suppose that six of the taxis are Checker Cabs and that the remaining four are Yellow Cabs. If the taxis are randomly dispatched, what is the probability that only Checker Cabs are dispatched to the first location?"
Am I to use permutations? Or is it combinations? Any help would be greatly appreciated.