# Permutation and combinations using chairs?

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.

• That depends on what exactly you mean by each of those questions. Are the people considered "different"? Are the chairs and their locations considered "different"? To what extent (e.g. are the chairs arranged in a circle and any rotation of the circle counts as the "same" arrangement or are they in a line?) When you say you want "$p$ people to be seated in $c$ chairs" what exactly does that mean? A unique pairing of persons and chairs? Do you intend for everyone to be seated? If so, what do you mean then when you have more people than chairs? – JMoravitz Jun 2 '18 at 18:50
• If you want as many people as possible to sit (but not necessarily all) then both of these will have $10\cdot 9\cdot 8\cdot 7\cdot 6$ outcomes. If you strictly want everyone to have a seat (and people can't share chairs) then there are zero valid arrangements for the second problem as there will always be some people without a chair. – JMoravitz Jun 2 '18 at 18:52

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.

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.

• For my answer, did I swap the cases or it is you? – Mr Pro Pop Jun 2 '18 at 19:52
• Sorry, don't understand what are you talking. – Aqua Jun 3 '18 at 6:33