The given question was: An airport bus drops off 35 passengers at 7 stops, each passenger is equally likely to get off to any stop and the passengers act independently of one another, the bus makes a stop only if someone wants to get off. Find the probability that the bus drops off passengers at ever stop.
I thought to proceed by stars and bars, basically, if x_1, x_2, ... , x_7 represent each of the stops the number of ways to x_1 + x_2 + ... + x_7 = 35 is (41 choose 6) since we add 6 bars/dividers to split the 35 numbers into 7 groups. Similarly to make sure one person drops at every station, we get x_1 + x_2 ... + x_7 = 28, since we subtract 7 to make sure every station has 1. This would give us (34 choose 6), using the similar starts and bars argument above.
Therefore the overall probability is (34 choose 6)/(41 choose 6).
Similarly we could solve this using PIE, by calculating 1 - P(A_1 U A_2 ... U A_7), where A_i represents the probability that no one stops at stop i.
Applying PIE, P(A_1 U A_2 ... U A_7) = 7*(6/7)^35 - (7 choose 2)(5/7)^35 + ... + (7 choose 7)(0/7)^35, which can we written as SUM from n = 1 to n =7, (7 choose n) * (-1)^(n+1) * ((7 - n)/(7))^35.
We then compute 1 - SUM to get the answer. I ran this sum into wolfram, and my stars and bars method, which gave differing answers. The sum gives around .97, whereas the stars and bars is .3. Can someone point me to something I am doing wrong. The PIE solution was the solution given to us by the textbook/professor.