A single ticket is needed for each possible pair of stations, since a passenger can travel from any station to any other station.

There are 5 stations, so we need to find the number of combinations of 2 stations that can be chosen from 5. This is a combination problem, and the formula for combinations is:

C(n, r) = n! / [r!(n-r)!]

where:
- n is the total number of options,
- r is the number of options chosen at a time,
- "!" denotes a factorial, which means multiplying all positive integers up to that number.

In this case, n = 5 (the total number of stations) and r = 2 (the number of stations involved in each ticket).

So, the calculation would be:

C(5, 2) = 5! / [2!(5-2)!]

= 5*4*3*2*1 / [2*1 * 3*2*1]

= 5*4 / 2

= 10

So, we need to prepare 10 kinds of single tickets.

Question

A single ticket is needed for each possible pair of stations, since a passenger can travel from any station to any other station.

There are 5 stations, so we need to find the number of combinations of 2 stations that can be chosen from 5. This is a combination problem, and the formula for combinations is:

C(n, r) = n! / [r!(n-r)!]

where:
- n is the total number of options,
- r is the number of options chosen at a time,
- "!" denotes a factorial, which means multiplying all positive integers up to that number.

In this case, n = 5 (the total number of stations) and r = 2 (the number of stations involved in each ticket).

So, the calculation would be:

C(5, 2) = 5! / [2!(5-2)!]

= 5*4*3*2*1 / [2*1 * 3*2*1]

= 5*4 / 2

= 10

So, we need to prepare 10 kinds of single tickets.

Knowee AI · Accepted Answer

A single ticket is needed for each possible pair of stations, since a passenger can travel from any station to any other station.

There are 5 stations, so we need to find the number of combinations of 2 stations that can be chosen from 5. This is a combination problem, and the formula for combinations is:

C(n, r) = n! / [r!(n-r)!]

where:
- n is the total number of options,
- r is the number of options chosen at a time,
- "!" denotes a factorial, which means multiplying all positive integers up to that number.

In this case, n = 5 (the total number of stations) and r = 2 (the number of stations involved in each ticket).

So, the calculation would be:

C(5, 2) = 5! / [2!(5-2)!]

= 5*4*3*2*1 / [2*1 * 3*2*1]

= 5*4 / 2

= 10

So, we need to prepare 10 kinds of single tickets.

5 stations on one line, how many kinds of single tickets do we need to prepare?

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