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There is an elevator with 8 people and 5 floors.With how may ways can these 8 people get out of the elevator,if we consider that they are not distinguishable?Which formula can I use to find it?

evinda
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1 Answers1

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Your question is the same as the following question :

Find the number of sets of $(a,b,c,d,e)\ (a,b,c,d,e\in\mathbb Z)$ such that $$a+b+c+d+e=8.$$

The answer is $$\frac{12!}{4!8!}=\binom{12}{4}=495.$$

Imagine you have 8 circles and 4 short lines, and arrange them in a row. Then, count the number of balls in each section divided by lines.

For example, the following represents we have $(a,b,c,d,e)=(2,2,0,1,3).$ And we can think this as that $2$ people get out of the elevator on the 1st floor, $2$ people on 2nd floor... $$\circ\circ|\circ\circ\ ||\circ|\circ\circ\circ$$

I hope this helps.

mathlove
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  • I understand...and,is there also an other way to calculate it? – evinda Dec 28 '13 at 15:13
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    What do you mean? Other way to calculate what? Your original question? or the example which I wrote? – mathlove Dec 28 '13 at 15:14
  • My original question..I understood your previous post,but I wanted to know if there is also an other way to find the number of ways the 8 people get out of the elevator. – evinda Dec 28 '13 at 15:18
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    Well, I suspect you don't understand my example well. My example is the same question as yours. Only difference is that I used circles and lines. Imagine that circles mean people, lines mean each floor. – mathlove Dec 28 '13 at 15:19
  • I understood your example..the general formula we get from this is $\binom{n+k-1}{k}$ .So,is there not an other way of thinking that could help me solve the exercise? – evinda Dec 28 '13 at 15:29
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    $\binom{n+k-1}{k-1}$, isn't? I don't know other ways though I don't know what kind of answers you want. Again, any other solution will be the same as mine, because mine is the same as yours. – mathlove Dec 28 '13 at 15:33
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    You are welcome! – mathlove Dec 29 '13 at 00:58