Number of ways of getting off the elevator

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SUMMARY

The discussion revolves around calculating the number of ways five people can exit an elevator on four floors of an apartment building. The correct calculation involves using combinations: $\binom{5}{0} \cdot \binom{5}{2} \cdot \binom{3}{2} \cdot \binom{1}{1}$, which simplifies to $\frac{5!}{2! \cdot 3!} \cdot \frac{3!}{2!} = 30$. This confirms that there are 30 distinct ways for the individuals to exit the elevator according to the specified distribution across the floors.

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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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