"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.
Well, we know the denominator of the fraction is 2520. You found that to be the total number of ways. Now all we need to know is how many ways can three checker cabs go the first location and the rest go to the other two locations.
Would that not be {combin(6,3)} (7!)/[(2!)(5!)]?
jettagrl,
Draw it out like this:
"C" = Checker
"Y" = Yellow
Location 1: CCC
Location 2: ***xx
Location 3: xx
Out of 6 Checker cabs, three of them end up at location 1, and there are 6C3 = 6! / (3!*(6-3)!) ways this can happen.
Now, location 1 is effectively "blocked out" from consideration, and we just concern ourselves with the remaining 7 slots. Within location 2 or 3, there can also be rearrangements that are identical for the purposes of this problem, so it's very similar to part a.
= 6C3 * [ 7! / (5!*2!) ]
Finally, divide this result by what you got in part a, which is the total number of ways that the cabs can be dispatched to the locations.
Interesting problem, John you and Plato both solved it as if the remaining 7 spots were in two sets of indistinguishable locations. i looked at it as 7 spots to be filled by any of seven taxis that are in two indistinguishable types which gives a slightly different result.
You have to l think this one over carefully to recognize that you are selecting spots and NOT taxis (which is why i was initially wrong), easy to confuse!
cheers
jerry
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