Question

In a three-storey building, there are four rooms on the ground floor, two on the first and two on the second floor. If the room are to be allotted to 6 persons, one person occupying one room only, the number of ways in which this can be done so that floor remains empty is
(A)${}^{8}{{P}_{6}}-\,2(6!)\,$
(B) ${}^{8}{{P}_{6}}$
(C)${}^{8}{{P}_{5}}\,(6!)\,$
(D) None of these

Answer 1

Firstly, we will find a total number of ways to arrange 8 rooms & 6 people using permutation. Then, there are cases where a floor may be empty and we will find the 2 cases where you can keep only the ${{1}^{st}}$and ${{2}^{nd}}$floor empty. Then we subtract it to get a number of ways.

Complete step by step answer:
Firstly, we have been given a three-storey building.
Now there are 4 rooms on the ground floor and 2 each on ${{1}^{st}}$and ${{2}^{nd}}$floor.
Then the total number of rooms in the 3-storey building can be given as:
Total number of rooms= rooms of ${{1}^{st}}$ floor + ${{2}^{nd}}$rooms of floor+ rooms of ground floor
$\text{2}+\text{2}+4=8$
Then, we get the total number of rooms as 8.
Now, these rooms are allotted to six persons in such a way that one person occupies one room only.
Thus, we get the number of allocations that can be done using Permutation.
Now, we use the definition of permutation as permutation of a set is an arrangement of its members into a sequence linear order, or if the set is already ordered, a rearrangement of its elements. Permutation is also the linear order of an ordered set. Thus, the number of permutation (ordered matters) of n things taken r at a time is given as:
${}^{n}{{P}_{r}}\,=\,\dfrac{n!}{(n-r)!}$
Thus from the above formula of permutation, we can say that the permutation of 8 rooms taken 6 at a time is given as${}^{8}{{P}_{6}}$, where n =8 and r= 6.
Among these, there would be cases where a floor may be empty.
Let’s say we can only keep the ${{1}^{st}}$ or${{2}^{nd}}$floor empty and then there are 2 cases In that case, the remaining 6 rooms are all occupied in${}^{6}{{P}_{6}}$ ways which is 6 rooms taken 6 at a time.
Thus we can say that the number of ways in which we can arrange 6 person, so that number floor remains empty$\text{=}\,\,{}^{8}{{P}_{6}}\,-\,2\times {}^{6}{{P}_{6}}$

& {}^{6}{{P}_{6}}\,\,=\,\,\dfrac{6!}{\left( 6-6 \right)!}\,\, \\\ & \Rightarrow {}^{6}{{P}_{6}}=\,\,6! \\\ \end{aligned}$$ So, we get the number of ways of arrangements as$${}^{8}{{P}_{6}}-2\left( 6! \right)$$. **So, the correct answer is “Option A”.** **Note:** To solve these types of questions we must have knowledge of basic concepts of permutation and factorial. Also, we should know that if 6 persons are to be arranged in 6 different rooms then the number of cases are 6!. So, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4. $\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$