Five friends go to a city where they can stay in 10 different hotels. In how many ways can they stay in these hotels if at least two of them stay together? Select the correct option
A) 5^10−10C5∗5! B) 5^10−10C5 C) 5^10−5! D) 10^5−10C5∗5! E) 10^5−10C5
There are 10 hotels and each person can stay in 10 hotels   Total = 10 * 10 * 10 * 10 * 10 --&gt; 10^5 --&gt; Notice that there is no restriction to avoid a hotel.   Number of ways so that no person stays together with another friend = 10 * 9 * 8 * 7 * 6 = 10P5 = 10C5 * 5!   Number of ways can 5 friends stay in these hotels if at least two of them stay together = 10^5 - (10C5 * 5!)   Answer: 10^5−10C5∗5!
<result>D</result>
Five friends go to a city where they can stay in 10 different hotels. In how many ways can they stay in these hotels if at least two of them stay together?