Question

How do I find the number of ways people can be seated on a bus?

30 people are to travel by bus. It can carry 17 passengers outside and 13 inside. 4 refuse to travel outside and 5 refuse to travel inside. How many ways can the passengers be seated?

Answer 1

`(13!)/(9!)(17!)/(12!)(21!)~~6.5xx10^29` ways for 30 people to sit on the bus

Explanation:

I'm going to assume that we're talking about an open-air double decker bus - one that has seating upstairs (with no roof) and downstairs (and so "inside").

We have a bus that will be full - we have 30 people who are going to travel in a bus that has 30 seats. If we had no other considerations, there would be `30! ~~ 2.7 xx 10^32` ways for people to sit.

However, we do have a couple of things to deal with. 4 people refuse to be outside (and so will be inside where there are 13 seats) and 5 people refuse to be inside (and so will be outside where there are 17 seats).

Let's work with these people first.

The 4 people who refuse to be outside will sit somewhere inside where there are 13 seats. This is a permutation problem, in that we do care in what seats they sit, and so we have:

`P_(n,k)=(n!)/((n-k)!); n="population", k="picks"`

`P_(13,4)=(13!)/(9!)=13xx12xx11xx10="17,160"` ways for these 4 people to sit.

We can do the same thing with the 5 people who refuse to sit inside - they'll be outside somewhere in the 17 available seats:

`P_(17,5)=(17!)/(12!)=17xx16xx15xx14xx13="742,560"`

And now we can work with the remaining passengers. There are 21 people who will sit anywhere on the bus, and so anywhere amongst the remaining 21 seats. That's `21! ~~ 5.1xx10^19`

In total then, we have:

`(13!)/(9!)(17!)/(12!)(21!)=13/(9!)(17!)(21!)~~6.5xx10^29` ways to sit people on the bus