MHB Number of ways of getting off the elevator

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Elevator
AI Thread Summary
The discussion revolves around calculating the number of ways five people can exit an elevator on four floors, with specific exit distributions: zero on the first floor, two on the second, two on the third, and one on the top floor. The calculation presented uses binomial coefficients to determine the combinations for each floor, resulting in a total of 30 ways. Participants confirm the accuracy of the calculation, expressing agreement with the method used. The conversation concludes positively, affirming the correctness of the solution. The final answer is indeed 30 ways for the specified exit distribution.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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:
 
Physics news on Phys.org
Looks good to me.
 
Ackbach said:
Looks good to me.

Nice,thank you! :)
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
Back
Top