Do I have to use n = 4 in the formula?

[evinda]

We have an elevator with 8 people and 5 floors.With how may ways can these 8 people get out of the elevator,when we know that at the first floor no one gets out?
I used the formula $$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=8$$ ,where $$x_{1}=0$$.So,it is $$\binom{n+k-1}{k}=\binom{4+8-1}{8}=\binom{11}{8}=495$$ ,right?Or do I have to replace n with 5?Because,the formula is satisfied for $$x_{i} \geq 0$$ ..

 

[I like Serena]

Re: Do I have to use n=4 at the formula?

We have an elevator with 8 people and 5 floors.With how may ways can these 8 people get out of the elevator,when we know that at the first floor no one gets out?
I used the formula $$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=8$$ ,where $$x_{1}=0$$.So,it is $$\binom{n+k-1}{k}=\binom{4+8-1}{8}=\binom{11}{8}=495$$ ,right?Or do I have to replace n with 5?Because,the formula is satisfied for $$x_{i} \geq 0$$ ..

You can reduce the problem to 8 people for which you pick 1 of 4 floors.
So $$x_{1}+x_{2}+x_{3}+x_{4}=8$$, meaning we indeed have $\binom{4+8-1}{8}$.
There is one problem though... $\binom{4+8-1}{8} \ne 495$...

 

[evinda]

Re: Do I have to use n=4 at the formula?

There is one problem though... $\binom{4+8-1}{8} \ne 495$...

Oh,yes..I am sorry! 495 was the result of an other subquestion and I wrote it accidentally.The result of the question I asked is 165 ;)

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You can reduce the problem to 8 people for which you pick 1 of 4 floors.
So $$x_{1}+x_{2}+x_{3}+x_{4}=8$$, meaning we indeed have $\binom{4+8-1}{8}$.

Thanks a lot!