/sci/ - Science & Math

[Anonymous No.10411228]

You have 3 cars, two cars cars are identical and interchangeable. You have 5 driveways. You can only park one car per driveway and call cars must be parked. How many different permutations can you have? Please explain how you solve this.

[Anonymous No.10411248]

>>10411228
60 permutations.
The first car has 5 possible options, any of the 5 driveways.
The second car has 4 possible options, once the first car is parked.
The third car, which is identical to the second, has 3 possible options once the second car is parked.

5*4*3=60

Really I don't think it matters that two of the cars are identical, but someone smarter than I am will probably correct me.

[Anonymous No.10411255]

>>10411248
for the identical case you have to just say
the second and third cars can be stuck into any of the four remaining driveways, so you have the options of remaining driveways 12, 23, 34, 13, 24, and 14, i.e. six ways to stick two identical cars in four driveways, so it's 5*6=30

it's half because there are double as many ways for the last two cars if they are distinguishable (as in your solution) -- you'd e.g. be allowed 12 and 21 instead of just 21 so it doubles all the ones i listed

[Anonymous No.10412548]

>>10411228
>You have 3 cars, two cars cars are identical and interchangeable. You have 5 driveways. You can only park one car per driveway and call cars must be parked.
none of those things are true