Welcome to our community

Be a part of something great, join today!

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

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
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 noone gets out?
I used the formula [tex] x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=8 [/tex] ,where [tex]x_{1}=0 [/tex].So,it is [tex] \binom{n+k-1}{k}=\binom{4+8-1}{8}=\binom{11}{8}=495 [/tex] ,right?Or do I have to replace n with 5?Because,the formula is satisfied for [tex] x_{i} \geq 0 [/tex] ..
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,791
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 noone gets out?
I used the formula [tex] x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=8 [/tex] ,where [tex]x_{1}=0 [/tex].So,it is [tex] \binom{n+k-1}{k}=\binom{4+8-1}{8}=\binom{11}{8}=495 [/tex] ,right?Or do I have to replace n with 5?Because,the formula is satisfied for [tex] x_{i} \geq 0 [/tex] ..
You can reduce the problem to 8 people for which you pick 1 of 4 floors.
So [tex] x_{1}+x_{2}+x_{3}+x_{4}=8 [/tex], meaning we indeed have $\binom{4+8-1}{8}$.
There is one problem though... $\binom{4+8-1}{8} \ne 495$... :eek:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: Do I have to use n=4 at the formula?

There is one problem though... $\binom{4+8-1}{8} \ne 495$... :eek:
Oh,yes..I am sorry!!! :eek: 495 was the result of an other subquestion and I wrote it accidentally.The result of the question I asked is 165 ;)

- - - Updated - - -

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