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

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The discussion revolves around calculating the number of ways 8 people can exit an elevator with 5 floors, given that no one exits on the first floor. The formula used is x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=8, with x_{1}=0, leading to the combinatorial expression \binom{n+k-1}{k}. It is clarified that the correct calculation involves \binom{4+8-1}{8}, which does not equal 495 as initially stated, but rather 165. The participants confirm that the problem can be simplified to choosing 1 of 4 floors for the exits.
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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 ..
 
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Re: Do I have to use n=4 at the formula?

evinda said:
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... $\binom{4+8-1}{8} \ne 495$... :eek:
 
Re: Do I have to use n=4 at the formula?

I like Serena said:
There is one problem though... $\binom{4+8-1}{8} \ne 495$... :eek:

Oh,yes..I am sorry! :o 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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I like Serena said:
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!
 
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