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vishal_kothari

  • 3 years ago

Seven people are in an elevator which stops at ten floors. In how many ways can they get o the elevator...

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  1. vishal_kothari
    • 3 years ago
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    get*off*the

  2. nubeer
    • 3 years ago
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    not sure maybe 10! x 7!

  3. vishal_kothari
    • 3 years ago
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    no..

  4. FoolForMath
    • 3 years ago
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    \( 11^7 \) ?

  5. vishal_kothari
    • 3 years ago
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    no..

  6. FoolForMath
    • 3 years ago
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    what is the answer ?

  7. FoolForMath
    • 3 years ago
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    @across: I don't think that's the right answer.

  8. vishal_kothari
    • 3 years ago
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    (a) 7^10 (b) 10^7

  9. vishal_kothari
    • 3 years ago
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    answer lies between this two...

  10. FoolForMath
    • 3 years ago
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    answer is \( 10^7 \) then.

  11. AnwarA
    • 3 years ago
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    Oh, I think the answer that across gave assumed that each one gets off in a different floor?

  12. vishal_kothari
    • 3 years ago
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    how?

  13. FoolForMath
    • 3 years ago
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    The problem is modeled as " how many ways can n distinct object can be divided in r distinct groups " some groups may be empty.

  14. FoolForMath
    • 3 years ago
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    My earlier answer assumes the super-set and I over counted few other cases.

  15. across
    • 3 years ago
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    FFM is correct.

  16. vishal_kothari
    • 3 years ago
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    ya..

  17. across
    • 3 years ago
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    Think of it in smaller terms: suppose there are three floors and two people; they can get off the elevator in 9 different ways, which is \(3^2\) as FFM's model states.\[\]

  18. FoolForMath
    • 3 years ago
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    Precisely, my earlier answer assumes that the seven people need not to get off the elevator at all and only some of them get down and all of the other obvious cases.

  19. FoolForMath
    • 3 years ago
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    @across: wanna try a variation " atleast one should get off in each floor" ? ; )

  20. vishal_kothari
    • 3 years ago
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    ok..

  21. across
    • 3 years ago
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    In this case, though, there are more floors than there are people. :p

  22. FoolForMath
    • 3 years ago
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    @vishal kothari: It's a bit hard, don't attempt it if you don't need it.

  23. FoolForMath
    • 3 years ago
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    @across: well just increase the numbers :P

  24. across
    • 3 years ago
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    Let's try 7 floors and 10 people.

  25. FoolForMath
    • 3 years ago
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    or more generally any \( r \gt n \)

  26. FoolForMath
    • 3 years ago
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    Sorry it should be: or more generally any \( r<n \) pertaining to my above model.

  27. across
    • 3 years ago
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    Whoops.

  28. FoolForMath
    • 3 years ago
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    @across: Sorry, that doesn't seem correct, as \( n^{ \underline{r} } \) is not the correct assumption, generally people attempt it with mutual inclusion-exclusion, but there is a even clever way, If you want I can give you a hint but it would be probably a spoiler.

  29. pokemon23
    • 3 years ago
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    anyone willing to explain me about square roots?

  30. FoolForMath
    • 3 years ago
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    Anyways, always excuse my solecism :P

  31. robtobey
    • 3 years ago
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    Assume that the elevator riders are considered indistinguishable, for example, 7 warm bodies. There are 11440 ways for them to get off of an elevator servicing 10 floors. http://2000clicks.com/mathhelp/CountingObjectsInBoxes.aspx Refer to: Indistinguishable Objects to Distinguishable Boxes The number of different ways to distribute n indistinguishable balls into k distinguishable boxes is C(n+k-1,k-1). For those who have access to Mathematica the following is a user defined function called Elevator where r is the number of riders and f is the number of floors. Elevator[ r_ , f_ ] := Binomial[ r + f - 1, f - 1]

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