Question

Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floors is
(a) $\frac{{}^{7}P_5}{7^5}$

(b) $\frac{7^5}{{}^{7}P_5}$

(c) $\frac{6}{{}^{6}P_5}$

(d) $\frac{{}^{5}P_5}{5^5}$

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)\frac{{}^{7}P_5}{{\left(7\right)}^5} \\ \text{Total possible ways of leaving the lift}=7\times 7\times 7\times 7\times 7={\left(7\right)}^5 \\ \text{5 people can leave different floors in}{}^{7}P_5\mathrm{ways}. \\ \mathrm{P}\left(5\text{people leaving the lift at different floors}\right)=\frac{{}^{7}P_5}{{\left(7\right)}^5} \end{gathered}$$