MHB Number of ways of getting off the elevator

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The discussion revolves around calculating the number of ways five people can exit an elevator on four floors, with specific exit distributions: zero on the first floor, two on the second, two on the third, and one on the top floor. The calculation presented uses binomial coefficients to determine the combinations for each floor, resulting in a total of 30 ways. Participants confirm the accuracy of the calculation, expressing agreement with the method used. The conversation concludes positively, affirming the correctness of the solution. The final answer is indeed 30 ways for the specified exit distribution.
evinda
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Hey! :)

An elevator of an apartment building with $4$ floors begins from the ground floor with $5$ people.Calculate the number of ways of getting off the elevator of $0$ zero people at the first floor , $2$ at the second floor, $2$ at the third floor and $1$ at the last floor.

I thought that it is:
$$\binom{5}{0} \cdot \binom{5}{2} \cdot \binom{3}{2} \cdot \binom{1}{1}=\frac{5!}{2! \cdot 3!} \cdot \frac{3!}{2!}=30$$

Could you tell me if it is right? :confused:
 
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Looks good to me.
 
Ackbach said:
Looks good to me.

Nice,thank you! :)
 
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