There are ten passengers on the train, and there are five stations along the way, so how many ways f

Isn't it just a matter of permutations and combinations? Understand?

It's a combination of permutations, and it's very simple

A!! Open the door to get off the train, and the person gets out of the train.

Six types of tickets are required for round trips. There are only a few trains with different round-trip fares for the same trip.

Now there is no problem of planning to prepare train tickets, and if you want to reimburse, you can also play "train tickets", and you need to play as many as you want.

Summary. There are 3 possibilities for the first passenger to get off the train (getting off at any of the 3 stations) and there are 3 possibilities for the second passenger to get off the train, so there are 3 3 = 9 possibilities for two passengers to get off the train.

There are 2 passengers on the bus, 3 stops along the way, and how many possible ways for passengers to get off the bus.

There are 3 possibilities for the first passenger to get off the train (getting off at any of the 3 stations), and there are also 3 possibilities for the second passenger to get off, so there are 3 3=9 possibilities for the two passengers to get off the train.

Can the principle of addition be used?

You can write out each of these situations one by one.

There are 3 ways to do this at each station.

There are 3+3+3 in 3 stations

OK. So what about 10 people getting off at 5 stations.

Multiplication or addition.

It's a combination problem. Assuming that each person can only get off at one hail station, the first person has 5 options (the drop-off station), the second person also has 5 options (can get off at the same station as the first person), and so on, and the tenth person also has 5 options. Therefore, the total number of drop-off scenarios is 5 5 5 5 = 5 10 = 9,765,625 possibilities.

The possible midsail properties are the same for each station, i.e. there are 5 possibilities for each station.

The column formula is: 42 13 + 7 = 29 + 7 = 36, and there are 36 passengers on the train before this station.

Summary. Hello, there are 6 passengers on the bus, there are 4 stops along the way, a total of 24 types.

There are 6 passengers on the bus and there are 4 stops along the way, so there are several ways to get off.

Hello, there are 6 passengers on the bus, there are 4 stops along the way, a total of 24 types.

According to the title, there are 4 possibilities for each passenger to get off the bus, 4+4+4+4+4+4=24.

Candidate station.

Suppose there are a, b, c, d, e, f, ......10 passengers in total. There are 5 stations A, B, C, D, and E.

Passengers A have 5 different options: A, B, C, D, and E, and the same ...... B, C, D, E, and FThere are also 5 different options for each. So the possible way is 5 5 5 5 5 ......, multiply 10 5s, which is 5 to the 10th power.

Since there are 992 passengers in total, and they sit in 8 cars on average, there are 992 8 = 124 people in each car.

You ask how many people are on a train with 8 cars and 118 cars per car? The train can accommodate about 944 passengers.

992 8 = 124, the answer is that each carriage did 124 people.

992 8 = 124, with an average of 124 people sitting in each carriage.

There were 124 people seated, and the column formula is 992 divided by 4 equals 124

10 people get off the bus at the station, 5 people get on the bus, and the number of people on the bus is 5. According to the relevant public information of the inquiry, 10 people get off the bus when the bus arrives at the station and 5 people get on the bus, so the remaining number of people on the bus is: 10 people - 5 people = (10-5 = 5) 5 people.

It's a matter of arrangement!

The first stop is to choose one person from 20 people to get off the train, and there are 20 ways to play;

The second stop is to choose one person from 19 people to get off, and there are 19 ways to get off;

The third station is to choose one person from 18 people to get off, and there are 18 ways to get off;

The fourth station is to choose one person from 17 people to get off, and there are 17 ways to get off;

At the fifth station, one of the 16 people can choose to get off the train, and there are 16 ways to get off.

So in total, there are 20 * 19 * 18 * 17 * 16 = 1860480 species.