A building has 10 doors and 5 persons will enter this building. What is the probability that the first four persons will enter the building through 4 different doors and the fifth person will use one of these 4 doors?
Choose 4 among 10 doors : P(10,4) ways
All possibilities :$\displaystyle 10^4$
Probability that the first 4 enter diff doors : $\displaystyle \frac{P(10,4)}{10^4}$
Probability that the fifth person enter the same door : $\displaystyle \frac{4}{10}$
Final answer :$\displaystyle \frac{P(10,4)}{10^4} \times \frac{4}{10}$
Is it correct?
The first person can chooses any of the 10 doors. The second person can choose any of the 9 remaining doors, the third 8 doors and the fourth 7 doors. The fifth person enter through any of the those four doors. There are, then 10(9)(8)(7)(4) ways the five people can enter. Without any restrictions there are $\displaystyle 10^5$ ways for 5 people to enter through 10 doors. The probability of the given situation is $\displaystyle \frac{10(9)(8)(7)(4)}{10^5}$. Since $\displaystyle _{10}P_4= \frac{10!}{(10- 4)!}= \frac{10!}{6!}= 10(9)(8)(7)$, that is the same as your answer.