MHB Solving Combinations Problem with up to 5 Roles: 40 Employees

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Combinations
AI Thread Summary
In a factory with 40 employees, the problem involves selecting a union of 5 people for 5 different roles, allowing employees to hold multiple roles. Since each of the 5 positions can be filled by any of the 40 employees, the calculation is based on sampling with replacement. Therefore, the total number of combinations is calculated as 40 raised to the power of 5, resulting in 40^5 possible unions. This approach differs from standard combination calculations, as it accounts for the possibility of the same employee occupying multiple roles. The final answer is 40^5 combinations for the union of 5 roles.
Yankel
Messages
390
Reaction score
0
Hello,

How do I solve this problem:

In a factory there are 40 employees. A union of 5 people is being chosen.
How many combinations are they, if the union of 5 people contains 5 different roles, and an employee can have more than one role (up to 5) ?

It is like sampling with replacement, so it isn't just:

{40 \choose 5}Thanks...
 
Physics news on Phys.org
Yankel said:
Hello,

How do I solve this problem:

In a factory there are 40 employees. A union of 5 people is being chosen.
How many combinations are they, if the union of 5 people contains 5 different roles, and an employee can have more than one role (up to 5) ?

It is like sampling with replacement, so it isn't just:

{40 \choose 5}Thanks...

It is not ${40 \choose 5}$..

There are $5$ positions and $40$ employees.

For the first position, there are $40$ possible employees.
For the second position, there are again $40$ possible employees, since the one that is chosen for the first position can also be chosen for the second position.
For the third position, there are again $40$ possible employees, for the same reason.
For the $4^{th}$ position, there are $40$ possible employees.
Fot the $5^{th}$ position, there are $40$ possible employees.

So, there are $\displaystyle{40^5}$ possible ways to create the union.
 
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.
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...
Back
Top