Question

Five people are going to be seated in a row of seven chairs. How many different ways can they be seated?

Answer 1

Hint: Here in the given question, we are asked to find the number of a possible arrangement of people in a collection of items which is chairs, given that the order of the selection does not matter. So, we find ways of selecting five chairs in a row of seven chairs in any order. Here we use the combinations formula to solve and then also represent the ways five people can be seated.

Complete step by step solution:
Given that Five people are going to be seated in a row of seven chairs.
To find the different ways they can be seated we use the combinations formula.
A combination is a mathematical technique that is used to find the number of different ways of possible arrangements in a group of items where the order of the selection does not matter.
It is given by the formula,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Where $C$ is the combination,
$n$ is the total number of objects and $r$ is the number of choosing objects in the set.
Here in our question,
$n$ will be $7$ which is the total number of objects and $r$ will be $5$ which is the number of choosing objects in the set.
We will be choosing $5$ chairs out of $7$ because there are five people.
Putting these values in the formula we get,
$\Rightarrow {}^{7}{{C}_{5}}=\dfrac{7!}{5!\left( 7-5 \right)!}$
On evaluating we get,
$\Rightarrow {}^{7}{{C}_{5}}=\dfrac{7!}{5!2!}$
On expanding the factorials, we get,
$\Rightarrow {}^{7}{{C}_{5}}=\dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\times 2\times 1}$
On simplifying by canceling the common terms we get,
$\Rightarrow {}^{7}{{C}_{5}}=\dfrac{7\times 6}{2}$
$\Rightarrow {}^{7}{{C}_{5}}=7\times 3$
$\Rightarrow {}^{7}{{C}_{5}}=21$
Now that there are $21\;$ ways to arrange the chairs,
And that there are five people, we represent them in different ways as $5!\;$
Therefore, the total number of ways is given by,
$\Rightarrow 21\times 5!$
On expanding the factorial, we get,
$\Rightarrow 21\times 5\times 4\times 3\times 2\times 1$
$\Rightarrow 2520$

Note: If at all there are two or more quantities in a permutation (who are the same of one kind) that are equal in number, then multiply the factorial of the number of kinds in the denominator also along with the other permutations.