Six cars pull up to a red light, one at a time. At the light, there are three lanes, one left-turn lane, one straight-going lane, and one right-turn lane. How many ways can the cars stack up so that all three lanes are occupied?
Note that if the first car turns left and the second goes straight, this is considered different from the first car going straight and the second car turning left. In other words, the cars are distinguishable, but pull up to the intersection in a fixed order.
Ok you didn't like the last answer.
(Incidentally why be so rude? If you feel the answer is incorrect respond back with why. Don't just silently downvote me)
So let's look at another way to count these.
There are 3 partitions of 6 using 3 numbers 1-4
(4,1,1), (3,2,1), (2,2,2)
These can be assorted among the lanes so that
(4,1,1) can be done 3 ways
(3,2,1) can be done 6 ways
(2,2,2) can be done 1 way
Now cars must be in order in the lanes.
(4,1,1) will have 6C4 * 2C1 = 30 arrangements. You pick 4 cars from 6, sort them, put them in their lane. Then select 1 car from the 2 remaining
and put it into a lane.
(3,2,1) will have 6C3 * 3C2 = 60 arrangements
(2,2,2) will have 6C2 * 4C2 = 90 arrangements
so there will be 3*30 + 6*60 + 1*90 = 90+360+90 = 540 arrangements
6 different from the last way I solved it. I'll have to figure out what's the disparity.