A fleet of ten taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and two go to airport C. a.In how many distinct ways can this be accomplished?b.Assume that taxis are allocated to airports at random. If exactly one of the taxis is in need of repair, what is the probability that it is dispatched to airport C?c.If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?

a) - Now, Three taxis can go to Airport A, 5 taxis can go to Airport B and 2 can go to Airport C so there are 10 taxis around. So, there are three groups, as well as the number of possible ways is as follows:

[tex]\therefore\;\;\;\;\;\;\;\;\;\; \frac{10!}{3!\times5!\times2!} = 2520[/tex]

b) - Although there is total 10 taxis, thus, there seems to be a possibility that such a car will be delivered to C airport is [tex]\frac{1}{10}[/tex].

c) - Therefore, when each of the groups will have precisely one hire car, it is necessary to repair it.

[tex]So,\;\;\;\;\;\;\;\;\;\;\; \frac{7!}{2!\times4!\times1!}=105[/tex]

Therefore, the method of assigning 3 repair taxis is 7, so the probability is as follows:

[tex]P=\frac{105\times7}{2520}[/tex]

[tex]=\frac{735}{2520} =0.2917[/tex]