How many ways are there and find the probability.

Question

A salesperson at a car dealership is showing cars to a prospective buyer. There are 9 models in the dealership. The customer wants to test-drive only 3 of them.

a. In how many ways could the 3 models be chosen if the order of test-driving is considered?

b. In how many ways could the 3 models be chosen if the order of test-driving is not important?

c. Suppose 6 of the models are new and the other 3 models are used. If the 3 cars to test-drive are randomly chosen, what is the probability that all 3 are new?

d. Is the answer to part (c) different depending on whether or not the order is considered?

My Work: a) P(9,3) = 504 ways

b) C(9,3) = 84 ways

c) ( C(6,3) x C(3,0) ) / C(9,3) = .2381

C(6,3) Choosing 3 new cars out of 6. C(3,0) Choosing 0 old cars out of 3, C(9,3) All possible combinations of the 9 cars.

d) No. replacing the above c) with P(6,3) and P(3,0) and P(9,3)

you get ( P(6,3) x P(3,0) ) / P(3,9) = .2381

You get the same answer as c.

My question is WHY? Why do you get the same answer as c? I thought the number of permutations is always greater then or equal to the number of combinations. Shouldn't this imply a greater probability if order is considered? Did I make a mistake somewhere? Is my work for a-d wrong? Any explanation for why d is the same as c would be greatly appreciated.

Answer 1

You could look at part D as: first pick out which 3 you want to test. The probability they are all new is what you worked out in part C.

After you have your cars picked out, you then decide on the order to drive them, which doesn't have any effect on the probability you got all new cars.

Hopefully this is at least a little bit helpful.

Answer 2

My question is WHY? Why do you get the same answer as c? I thought the number of permutations is always greater then or equal to the number of combinations.

Indeed that is true; yet, here you have both the numerator and denominator being coincidentally larger by the same factor ($3!$), and that common factor cancels.

$\dfrac{\mathrm C(6,3)~\mathrm C(3,0)}{\mathrm C(9,3)}=\dfrac{\dfrac{6!}{3!~3!}\dfrac{3!}{0!~3!}}{\dfrac{9!}{3!~6!}}=\dfrac{6!~6!}{9!~3!}$

$\dfrac{\mathrm P(6,3)~\mathrm P(3,0)}{\mathrm P(9,3)}=\dfrac{\dfrac{6!}{3!}\dfrac{3!}{3!}}{\dfrac{9!}{6!}}=\dfrac{6!~6!}{9!~3!}$

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However: You actually want $\tfrac{\mathrm P(6,3)~\mathrm P(3,0)~\mathrm C(3,0)}{\mathrm P(9,3)}$, since you are considering the order of cars selected, so must count ways to queue up three from six new cars, queue up zero from three old cars, and then ways to splice those two queues together into one.

Which is numerically the same answer because the extra factor is one -- yet the logic is important.   Consider an event where the splicing of queues is not trivial: the probability for testing two new and one old model is $$\dfrac{\mathrm P(6,2)~\mathrm P(3,1)~\mathrm C(3,2)}{\mathrm P(9,3)}$$