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If 4 people have 5 different cars to choose from and two of those people cannot pick the same(the remaining two people could have the same car). How many different ways could people pick the cars?

At first I was thinking

First Person: 5 choices Second Person: 4 choices, because they cannot have the same car as person 1. Third Person 5 choices Fourth Person 5 choices Overall 5*4*5*5 = 500 ways, but I'm just not sure if this would assure people 1 and 2 don't get the same car.

To help with confusion lets just give the people names to make this more clear. Abby, Bob, Chris, and Dan. They have 5 different types of cars to choose from but Abby and Bob cannot have the same type of car. Meaning Abby, Chris, and Dan could have the same type of car or Bob Chris and Dan could have the same car. The only restriction is that Abby and Bob do not have the same type of car

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  • $\begingroup$ You were so close... but the third person cannot pick what P1 picked, or what P2 picked! So there are three choices for P3. (Obviously, amWhy's answer covers all this.) $\endgroup$ – The Chaz 2.0 Oct 16 '13 at 17:58
  • $\begingroup$ What do you want to say by "two people cannot pick the same"? A) "only two people cannot pick the same" or B) "any two people cannot pick the same"? $\endgroup$ – Cristi Oct 16 '13 at 18:11
  • $\begingroup$ Only two people cannot pick the same. $\endgroup$ – Nate Oct 16 '13 at 18:17
  • $\begingroup$ If first person has 5 choices, second has also 5, but third has already only 3 or 4 choices depending on the first two persons pick; forth has also 2 or 3 choices left, depending on the first three persons pick... $\endgroup$ – Cristi Oct 16 '13 at 18:19
  • $\begingroup$ So lets just give the people names to make this more clear. Abby, Bob, Chris, and Dan. They have 5 different types of cars to choose from but Abby and Bob cannot have the same type of car. Meaning Abby, Chris, and Dan could have the same type of car or Bob Chris and Dan could have the same car. The only restriction is that Abby and Bob do not have the same type of car. $\endgroup$ – Nate Oct 16 '13 at 18:24
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If no one can have the same car, and/or no pair of people can have the same care, then we have $$\;5!\;= \;5\cdot 4\cdot 3\cdot 2 = 120$$ ways, since no one can have a car the same as anyone else.

You logic was correct about person $2$ not having as a choice the car selected by Person $1$. By the same logic, Person $3$ cannot have as a choice either the car selected by Person $1$, nor the car selected by Person $2$, leaving Person $3$ only 3 choices, and so on.


Given your clarification, then we have that the only restriction is that Person 1 and Person 2 do not have the same car, then your calculation $5\cdot 4\cdot 5\cdot 5$ is correct.

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  • $\begingroup$ I don't think your answer is correct $\endgroup$ – Cristi Oct 16 '13 at 17:56
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    $\begingroup$ @Cristi: I do think her answer is correct! $\endgroup$ – The Chaz 2.0 Oct 16 '13 at 17:57
  • $\begingroup$ If first person has 5 choices, second has 5, but third has already only 3 or 4 choices depending on the first two persons pick; forth has also 2 or 3 choices left, depending on the first three persons pick... $\endgroup$ – Cristi Oct 16 '13 at 18:00
  • $\begingroup$ You're wrong, @Cristi. Person 1 and 2 cannot pick the same car, which might happen if has 5 choices. $\endgroup$ – Namaste Oct 16 '13 at 18:01
  • $\begingroup$ It just says two people cannot pick up the same, not the first 2. Maybe I am wrong... $\endgroup$ – Cristi Oct 16 '13 at 18:02

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