Question

In how many ways can five people be arranged in three different rooms if no room must be empty and each room has 5 seats in a single row.

Answer 1

Here, we will use the permutation and combination method to find the required answer. First, we will find the number of basic possibilities in which these 5 people can be arranged in 3 rooms. Then we will use the permutation method to find the required number of ways to arrange these people in 3 different rooms for these possibilities individually and then we will add the obtained values to get the final answer.

Complete step by step solution:
Here we need to find the number of ways to arrange five people in three different rooms so that no room is empty.
There are 2 basic possibilities:-
$\left( {1,2,2} \right)$
$\left( {1,1,3} \right)$
We will first consider case 1.
It is given that there are five seats in a room.
Therefore, the number of ways to arrange these people $= \dfrac{{5!}}{{1! \times 2! \times 2! \times 2!}} \times {}^5{C_1} \times {}^5{C_2} \times 2! \times {}^5{C_2} \times 2!$
We know that the formula of combination is
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
Now, we will use this formula to calculate the required value;
Number of ways to arrange these people $= 15 \times 3! \times \dfrac{{5!}}{{\left( {5 - 1} \right)! \times 1!}} \times \dfrac{{5!}}{{\left( {5 - 2} \right)! \times 2!}} \times 2!\dfrac{{5!}}{{\left( {5 - 2} \right)! \times 2!}} \times 2!$
On simplifying the terms, we get
Number of ways to arrange these people $= 15 \times 3! \times \dfrac{{5!}}{{4!}} \times \dfrac{{5!}}{{3! \times 2!}} \times 2!\dfrac{{5!}}{{3! \times 2!}} \times 2!$
On finding the value of factorials, we get
Number of ways to arrange these people $= 15 \times 3 \times 2 \times 1 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2}} \times 2 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2}} \times 2$
On further simplification, we get
Number of ways to arrange these people $= 15 \times 3 \times 2 \times 5 \times 5 \times 2 \times 2 \times 5 \times 2 \times 2$
On multiplying the numbers, we get
Number of ways to arrange these people $= 1,80,000$
Now, we will consider case 2.
It is given that there are five seats in a room.
Therefore, the number of ways to arrange these people $= \dfrac{{5!}}{{3! \times 1! \times 1! \times 2!}} \times {}^5{C_1} \times {}^5{C_1} \times {}^5{C_3} \times 3!$
We know that the formula of combination is ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ .
Now, we will use this formula to calculate the required value.
Number of ways to arrange these people $= 10 \times 3! \times \dfrac{{5!}}{{\left( {5 - 1} \right)! \times 1!}} \times \dfrac{{5!}}{{\left( {5 - 1} \right)! \times 1!}} \times \dfrac{{5!}}{{\left( {5 - 3} \right)! \times 3!}} \times 3!$
On simplifying the terms, we get
Number of ways to arrange these people $= 10 \times 3! \times \dfrac{{5!}}{{4! \times 1!}} \times \dfrac{{5!}}{{4! \times 1!}} \times \dfrac{{5!}}{{2! \times 3!}} \times 3!$
On finding the value of factorials, we get
Number of ways to arrange these people $= 10 \times 3 \times 2 \times 1 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2}} \times 3 \times 2$
On further simplification, we get
Number of ways to arrange these people $= 10 \times 3 \times 2 \times 5 \times 5 \times 5 \times 2 \times 3 \times 2$
On multiplying the numbers, we get
Number of ways to arrange these people $= 90,000$

Hence, the total number of ways to arrange people in three rooms $= 1,80,000 + 90,000 = 2,70,000$

Note:
Here we have calculated that factorial. So, we need to keep in mind following properties of factorial:

1. Factorial of any positive integer can be defined as the multiplication of all the positive integers less than or equal to the given positive integers.
2. Factorial of zero is one.
3. Factorials are generally used in permutations and combinations problems.
4. Factorials of negative integers are not defined.