Question

In how many ways four friends can put up in $8$ hotels of a town if
$\left(i\right)$ No two friends can stay together
$\left(ii\right)$ All the friends do not stay in the same hotel?

• A. $\left(i\right)1680\left(ii\right)4088$
• B. $\left(i\right)1680\left(ii\right)3670$
• C. $\left(i\right)1540\left(ii\right)4088$
• D. $\left(i\right)1540\left(ii\right)3670$

Answer 1

The correct option is A $\left(i\right)1680\left(ii\right)4088$

(i) The first has 8 hotels options, second has 7 options...
So total $=\mathrm{8.7.6.5}=1680$ ways
(ii) If there are no such restriction they can be allotted hotels in $\mathrm{8.8.8.8}$ ways = $4096$ ways
We have to rule out possibility where all of them are in the same hotel, it can happen in 8 ways (as there are 8 hotels)
So total $=4096-8=4088$ ways
Hence, option 'A' is correct.