Hello, peter_home!

There are 10 passengers on a bus.

The bus makes three stops: A, B, C.

A) In how many ways can they can get off the bus?

For each of the 10 passengers, there are 3 choices of stops.

The number of ways is: .

V) In how many ways they can get off from the bus

if there must be 4 at any two stops and 2 at the other?

There are three distributions: .

For any one of them, there are: . ways.

Therefore, there are: . ways.

G) In how many ways they can get off the bus

if 4 must get off at A or 4 must get off at B (but not both)?

I will assume this means "exactly4 get off at A" or "exactly4 get off at B."

Four get off at A

There are: . ways for four to get off at A.

The other six must choose between B and C: . ways.

. . But this includes the cases where four get off at B.

. . How many are there?

. . . . If four get off at B, there are: . ways

. . . . The remaining two get off at C: . way.

. . . . So, there are 15 cases in which four also get off at B.

So, there are: . ways for the other six passengers.

Hence, there are: . ways for four to get off at A.

By similar reasoning, there are . ways for four to get off at B.

Therefore, there are: . ways.

Someone check my reasoning and my work . . .please!

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