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Homework Help: How to find r in the equation P(5,r)=20

  1. May 5, 2005 #1
    how would you solve for r in this situation:

    P(5, r) = 20

    Now I understand you cant use trial and error to find that r=2 but is there any other way using an algebraic technique. Itried something like this..
    [tex] \frac{5!}{(5-r)!}=20[/tex]

    now i don't know what to do next
  2. jcsd
  3. May 5, 2005 #2
    Express 6 as a factorial.
  4. May 5, 2005 #3
    And consider what [tex](5-r)![/tex] means.

    [tex](5-r)! = (5-r)*(5-r-1)*(5-r-2)[/tex] etc...

    That with whozum's help should help you see the cancellation (hint).
  5. May 5, 2005 #4
    ahh i see then you get 3!=(5-r)!, then cancel out the factorials and you get 3=5-r.. solve for r .. r=2 .. thanks, thats been bothering me alot

    i got another question but im not sure if it has to do with permutations or but it has a similar subject. should i post it in this topic?
  6. May 5, 2005 #5
    Go ahead. Also I didnt check your work on the first one, but if all is correct til the last step then r = 2.
  7. May 5, 2005 #6
    3 Canadians, 4 Americans, and 2 Mexicans attend a trade conference.

    a) In how many ways can they be sated in a row if people of the same nationality are to be seated next to eachother?

    b) redo (a) if they sit at a round table

    for a) the way i look at it is for the canadians there are three ways which they can be seated so you get 3!, then same for the americans and mexicans getting 4! and 2! respectively. Then you have a total of three nationalities which means they can be switched around giving you 3!.. Therefore in total you get


    not sure if thats right

    b)im pretty sure this is wrong but heres what im thinking
  8. May 5, 2005 #7
    any takers?
  9. May 5, 2005 #8
    k scratch what i did for (b) i just figured out thats wrong.. assuming what i did in (a) was right, then (b) should be something like
    [tex]1\bullet2!(1\bullet2!+1\bullet3! +1\bullet1!)[/tex]
    [tex]=2(9) = 18[/tex]
  10. May 6, 2005 #9


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    {# in C Block} = 3
    {# in A Block} = 4
    {# in M Block} = 2

    There are (3!) arrangements of the 3 Block Units. For each Block Unit arrangement, there are (3!) sub-arrangements within the "C" Block, (4!) sub-arrangements within the "A" Block, and (2!) sub-arrangements within the "M" Block.
    {Total Arrangements} = (3!)*{(3!)*(4!)*(2!)}

    The round table has 9 chairs. Let the chair closest to North direction be designated "N". Chair "N" will be occupied by a member of EITHER "C", "A", or "M". If it's "C", there are 3 possible Block positions relative to "N" (0 members right of "N", 1 member right of "N", & 2 members right of "N"). For each relative position, there are (3!) sub-arrangements. Further, the 2 remaining Blocks ("A" & "M") have 2 possible relative positions (to left of "N" or to right of "N"), with (4!) sub-arrangements within "A" Block and (2!) sub-arrangements within "M" Block. Thus, total arrangements possible when "C" Block occupies chair "N" is:
    {(3)*(3!)}*(2)*{(4!)*(2!)} ::: <--- "C" occupies "N"
    Calculate similar results for cases when Blocks "A" & "M" occupy chair "N", and add all three results together for total number of arrangements:
    {(3)*(3!)}*(2)*{(4!)*(2!)} + ????? + ?????

  11. May 6, 2005 #10
    so for b) you get

    [3(3!)][2(4!2!)] + [2(2!)][2(3!4!] + [4(4!)][2(3!2!)]

    Alright thanks
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