Question

In a three-story building, there are four rooms on the ground floor, two on the first and two on the second floor. If the rooms are to be allotted to six persons, one person occupying one room only, the number of ways in which this can be done so that no floor remains empty is-
A. ${}^8{P_6}$- 2 (6!)
B. ${}^8{P_6}$
C. ${}^8{P_6}$(6!)
D. None of these

Answer 1

Hint- In this question we use the theory of permutation. So, before solving this question you need to first recall the basics of this topic. For example, if we need to arrange two persons in the four rooms. Then in this case, this can be done in${}^{\text{4}}{P_2}$ .
Formula used- ${}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[(n - r)!]}}}}$
Total number of allocation methods = ${}^8{P_6}$

Complete step-by-step answer:
Among these, there would be cases where a floor may be empty. We can only keep the 1st or 2nd floors empty. Thus, there are 2 cases.
In that case, the remaining 6 rooms are all occupied in${}^6{P_6}$ways.
We have total 8 rooms
Thus, final probability = ${}^8{P_6}$- 2 (${}^6{P_6}$)
= ${}^8{P_6}$- 2 (6!)
Thus, the answer is ${}^8{P_6}$- 2 (6!)
Therefore, ${}^8{P_6}$- 2 (6!) ways in which we can be done so that no floor remains empty in the given circumstances.
Thus, option (A) is the correct answer.

Note- A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. For example, suppose we have a set of
three letters: A, B, and C. Each possible selection would be an example of a combination.