# Math Help - Passengers...

1. ## Passengers...

Good morning to all

10 passengers that are considered distinguished, find itself in a bus and remain 3 attitudes (StA, StB, StC)

A.With how many ways they can get off from the bus all the passengers if in each bus stop it's possible to get off from no one up to 10 passengers?
V. With how many ways they can get off from the bus all the passangers if they have to get off 4 in two any bus stop and 2 in the other?
G. With how many ways they can get off all of the bus if they have to get off 4 in the StA or 4 in the StB( however they can not get off 4 in the two simultaneously)?

In all the questions will not be taken into consideration the order of geeting off the passangers from the bus

2. 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: . $3^{10} \:=\:\boxed{59,049}$

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: . $(A,B,C) \:=\:(4,4,2),\:(4,2,4),\:(2,4,4)$

For any one of them, there are: . ${10\choose4,4,2} \:=\:3150$ ways.

Therefore, there are: . $3 \times 3150 \:=\: \boxed{9450}$ 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 "exactly 4 get off at A" or "exactly 4 get off at B."

Four get off at A

There are: . ${10\choose4}\,=\,{\color{blue}210}$ ways for four to get off at A.

The other six must choose between B and C: . $2^6 = 64$ ways.

. . But this includes the cases where four get off at B.
. . How many are there?
. . . . If four get off at B, there are: . ${6\choose4} = 15$ ways
. . . . The remaining two get off at C: . $1$ way.
. . . . So, there are 15 cases in which four also get off at B.

So, there are: . $64 - 15 \:=\:{\color{blue}49}$ ways for the other six passengers.

Hence, there are: . $210 \times 49 \:=\:10,290$ ways for four to get off at A.

By similar reasoning, there are . $10,290$ ways for four to get off at B.

Therefore, there are: . $10,290 + 10,290 \:=\:\boxed{20,580}$ ways.

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

3. The last one does ckeck.
$2{{10}\choose 4} \left( {\sum\limits_{k = 0}^6 {\left( {k \ne 4} \right){{6}\choose k}} } \right) = 20580
$