Permutation and combinations using chairs?

Question

After reading and watching a lot about permutation, combination and variation i still don't understand them fully. So i have two questions:

1. How many ways are there to position 5 people on 10 chairs?
2. How many wars are there to position 10 people on 5 chairs?

Please include solving so i can learn from it.

Answer 1

How many ways are there to position $5$ people on $10$ chairs?

We have 10 chairs and $5$ people to be seated. So the answer is ${ 10 \choose 5}$ to select which chairs people will be seated on. If the order matters, you will need to multiply that by $5!$ which is the number of ways that people can be arranged among themselves.

$$ \binom{10}{5} \times 5! = 252 \times 120 = 30240$$

Here $\binom{n}{k}$ is the binomial coefficient which denotes the number of ways to choose $k$ objects from a collection of $n$ distinct objects.

The number $\binom{n}{k} \times k!$ is also denoted as $^nP_k$.

How many wars are there to position $10$ people on $5$ chairs?

We have $5$ empty slots where people could be seated __ __ __ __ __ For the first one, any of the $10$ people could seat, then $9$ people are left to be seated on the second chair, the third now has $8$ choices, leaving $7$ choices for the fourth chair, and finally $6$ for the fifth chair.

By the fundamental principle of counting, there are a total of $10*9*8*7*6 = 30240$ different seating arrangements.

Answer 2

1. First man has $10$ choises, then second one $9$ and so on... we get $10\cdot 9\cdot 8\cdot 7 \cdot 6$ ways to arrange $5$ people if we have $10$ chairs.

2. We first choose $5$ people among $10$ which are going to be sited, this we can do on ${10\choose 5}$ ways, then we arrange this 5 people on $5!$ ways. So we have $${10\choose 5}\cdot 5!$$ ways $10$ people to be sited.