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Math Help - Permutattion problem

  1. #1
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    Permutattion problem

    A plane flight has been cancelled and the 28 passengers must be booked onto 4 alternate flights called flights A,B,C,D with each subject to the following restrictions:

    i) Flight A has 12 empty seats but only 9 are available to passengers from the cancelled flight.

    ii) Flight B has 8 empty seats, all available to passengers from the cancelled flight.

    iii) Flight C has 14 empty seats, 5 of which are available to passengers from the canceled flight.

    iv) Flight D has 9 empty seats, six of which are available to passengers from the canceled flight.

    Passengers as well as seats are distinct

    a.) How many different ways can passengers from the canceled flight be assigned seats on flight B?

    b.) How many different ways can passengers from the canceled flight be assigned seats on flights A and B?

    c.) How many ways can the passengers from the cancelled flight be assigned to flights A,B,C?

    d.) How many ways can passengers from the canceled flight be assigned to flights A,B,C,D?
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  2. #2
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    This is truly a simplistic probem, assuming all seats are different.
    P(n,k)=\frac {n!} {(n-k)!}
    a) P(28,8)
    b) P(28,9+8)
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  3. #3
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    How about:

    a.) First of all we need to choose which 8 lucky people of the 28 get to go on flight B. There are C(28,8) such choices. Now for each such choice there are 8 seats avl. for the first person, 7 for the second, 6 for the third, and so on. Leading to a total number of possibilities C(28,8)8!

    b.) We need to choose 8 people to go on flight B,and 9 people to go on flight A. C(28,8)C(20,9) such possibilities. Now of the 8 chosen for flight B there are 8! seating arrangements. For flight A it is more complicated, we need to choose 9 of the 12 available seats for the inconvenienced passengers. So there are C(12,9)9! choices for the flight A people's seating arrangements. In total then there are C(28,8)C(20,9)C(12,9)9!8!.

    Others are similar.
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  4. #4
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    I am still not quite sure

    Is there anyway you can break it down into steps I would take to compute each one!!!
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  5. #5
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    Quote Originally Posted by digitalis77 View Post
    Is there anyway you can break it down into steps I would take to compute each one!!
    You don't think that I did that for you?
    If you don't, the I conclude that you simply want the answers.
    AND do not want to do them for yourself.
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  6. #6
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    Sorry I will take another look!!

    Sorry I will take another look!!
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  7. #7
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    If you apply Plato's formula you will derive the same values I posted for problems a.) and b.) with the exception that I am assuming the 9/12 seats in flight A are not fixed. The formula for permutations is well known but may not appear immediately applicable to a given problem. I also have given what I think to be a detailed and explicit explanation for the reason you use the permutation function for these problems. It follows from the cancellation of a term in the denominator of the choose function by multiplication of the number of choices by the number of arrangements. C(n,r) = n!/r!(n-r)! P(n,r) = n!/(n-r)! = C(n,r)r!.
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  8. #8
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    How would this one look with Platos formula? If it was written out fraction style?

    b) P(28,9+8)
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  9. #9
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    My answer for b.)

    So here is my answer for b.) 28C17 X 12C3 X 17P17

    What do you think, does it work?
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  10. #10
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    For b.) P(28,17) = \frac{28!}{(28-17)!} = \frac{28!}{11!} This comes from the description I gave, but assumes a fixed number of seats (9) are available to the 9 passengers going to flight A in the following way:

    Of the 28 people first choose 8 to go to flight B,  C(28,8) = \frac{28!}{8!(28-8)!} Then arrange those 8 people however you want, 8! such arrangements: \frac{28!}{8!(28-8)!}8! = \frac{28!}{(20)!}.

    Now choose 9 of the remaining 20 people to go on flight A and arrange them in every possible way:  C(20,9)9! = \frac{20!}{9!(20-9!)}9! = \frac{20!}{11!}.

    So the total number of possibilities is a product of the above two results: \frac{28!}{(20)!}\frac{20!}{(11)!} = \frac{28!}{11!} = \frac{28!}{(28-17)!} = P(28,17) .

    However I assumed that the 9 seats available to the 12 passangers were not fixed. In other words there were 9 seats available to the displaced passangers and any of the 12 vacant seats could be filled by the 9 passengers on flight A. Perhaps the scenario is exactly 3 seats must be left vacant because of weight restrictions. Then obviously it doesn't matter which 9 seats are chosen of the 12, so there is the additional choice of 9 out of 12 seats which must be counted - leading to a final answer of P(28,17)C(12,9).

    Make sense?
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  11. #11
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    Yes, it makes sense.

    I just am having a hard time understanding the symbolism you use to write. My answer I came up with is 12C9 X 28C9 X 9! X 19C8 X 8!

    Can you see the way I do things. Let me know.
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  12. #12
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    Yes:
    C(n,r), if you're writing in "standard form"

    is

    nCr, (if you're on your calculator)

    is

    n choose r, (if you're using Google calc)

    is

    n!/r!(n-r)!, (if you care to be explicit about the formula)

    is

     n \choose r , (if you can write in LaTex)

    is

    C_n^k , (if... I don't know what)

    is

    The number of subsets of size r in a set of size n. (If you like counting)
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