In how many ways can a cinema hall with 100 seats be filled with 100 people such that certain conditions are satisfed?
Conditions:
- 40 seats are on the first floor
- 60 seats are on the ground floor
- 10 of the 100 people will only be seated on the first floor and 15 others will only be seated on the ground floor
My solution:
Select 10 seats from the first floor and find the arrangements for the 10 people ( $^{40}C_{10} \times10!$ ), then similarly select 15 seats from the ground floor and find arrangements for the 15 people ($^{60}C_{15}\times15!$), and then arrange the rest in the remaining seats(75!): $$^{40}C_{10} \times10!\times^{60}C_{15}\times15!\times75!$$
But this is incorrect, why is this incorrect and how to solve this?
Best Answer
Your answer is correct.
Here is another approach to the problem:
We know that we must reserve $10$ seats on the first floor for the $10$ people who will only sit on the first floor and $15$ seats on the ground floor for the $15$ people who will only sit on the ground floor. Therefore, there are $40 - 10 = 30$ seats left on the first floor. We must choose which $30$ of the remaining $100 - 10 - 15 = 75$ people will sit on the first floor. We can do this in $\binom{75}{30}$ ways. The remaining $75 - 30 = 45$ people must take the remaining $60 - 15 = 45$ seats on the ground floor. The $40$ people on the first floor can be seated in the $40$ seats on the ground floor in $40!$ ways. The $60$ people on the ground floor can be seated in the $60$ seats on the ground floor in $60!$ ways. Hence, there are $$\binom{75}{30}40!60!$$ admissible seating arrangements, which agrees with your answer.
The stated answer $$\binom{75}{30} = \frac{75!}{30!45!}$$ is the number of ways we can select which $30$ of the $75$ people who have not indicated that they will only sit on a particular floor will be seated on the first floor. It is not the answer to the question you stated.