24 people in 3 different hotels

Question

If I were to place 24 people in three different hotels (8 people at each hotel). How many ways are there to do this?

First and foremost I understand that this is a combination problem, (having a hard time adjusting to permutation and combinations but have finally understood that permutation is when the order counts)

So how would I go on solving this problem? Am I doing this correct?

24 C 8 = 24!/(8! x !16) = 735471 ways?

Answer 1

First you need to choose $8$ from the $24$ people to stay in the first hotel and then you choose $8$ from the remaining $16$ for staying in the $2$nd hotel. The tenants of the $3$rd hotel are fixed by the first two choices.

So, total number of such ways will be $\binom{24}{8} \times \binom{16}{8} = \frac{24!}{(8!)^3}=9465511770$.