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anonymous

  • 5 years ago

Find the number of different arrangements of the letters in each word. MANHATTAN if the word MAN must appear in the arrangement

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    7!/(2!*2!)

  3. anonymous
    • 5 years ago
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    is that so?no more conditions?

  4. anonymous
    • 5 years ago
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    my attempt at this was (1*3*2) * 6! / (3! * 2!)

  5. anonymous
    • 5 years ago
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    note:the dividing factorials are for the repitions of individual letters in the arrabgement

  6. anonymous
    • 5 years ago
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    where do you get the 7! from?

  7. anonymous
    • 5 years ago
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    take "MAN" as one unit (letter ) since we know we can't break that up .i.e it must exist in the final answer then the answer is simply the number permutations. correction:7!/(3!*2!*2!)

  8. anonymous
    • 5 years ago
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    u can arrange 7 letters in 7! ways, the 3 letters of MAN in 3! ways and since arrangements of TT and AA are same so divide by 2!*2! But since MAN should appear as arranged ie A after M and N after A...so we take only one arrangement..and so the answer is 7!/(2!*2!)

  9. anonymous
    • 5 years ago
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    no, the relative permutation of "man" cannot be changed.

  10. anonymous
    • 5 years ago
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    but there are several possibilities of A and N in the M.A.N. combination so wouldn't it be 1*3*2 * (6! [the rest of the letters including the extra A's and N's)

  11. anonymous
    • 5 years ago
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    that fact leads to the dividing factorials ,but the 3! comes from the fact that there are 3 a's and not that there are 3 alphabets in "man"

  12. anonymous
    • 5 years ago
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    druvidfae is wrong..

  13. anonymous
    • 5 years ago
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    i just don't get why MAN is assumed to be 1 letter there's 1 possibly for M and then 3 a's and 2n's so can't it be 1M 1A 1N 1M 1A 2N etc. why isn't it (1*3*2) for the individual possibilities of MAN combinations

  14. anonymous
    • 5 years ago
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    see u need MAN to be dere in any arrangement..so it will help if we take it as one letter.

  15. anonymous
    • 5 years ago
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    so how would you find the number of different arrangements with no restrictions and given the problem: MANHATTAN?

  16. anonymous
    • 5 years ago
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    u need MAN to appear as M-A-N! lol :D I mean u r not allowing ANM or MNA or NAM

  17. anonymous
    • 5 years ago
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    9!/(2!3!2!)

  18. anonymous
    • 5 years ago
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    i replied tat for no restrictions case..

  19. anonymous
    • 5 years ago
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    so then with this restriction it would be... 7!/(3!2!2!) with 3! accounting for a's, 2! accounting for n's and t's?

  20. anonymous
    • 5 years ago
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    no when u r counting with the given restriction u cant permute the letters of MAN..understood? Use it as another letter which is not there in thee word..like Z..so ur given problem becomes: permute Zhattan

  21. anonymous
    • 5 years ago
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    U cant play with the letters of MAN since they need to be arranged..when u account for them as u said then MAN will not appear as MAN in all ur arrangements...

  22. anonymous
    • 5 years ago
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    oh okay that makes sense when you change it to a different letter

  23. anonymous
    • 5 years ago
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    yah..but u must understand the deep meaning also...why we change it to a different letter!

  24. anonymous
    • 5 years ago
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    Cause when account for a's and n's of MAN (which u replied in a post) then u get some arrangements as: MANHATTAN NAMHATTAN HATTAMNAN.... but u want MAN as M-A-N not other arrangements..so u keep that fixed and take Z

  25. anonymous
    • 5 years ago
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    because you want the combination M A N to be there in any arrangement so you should assume its one letter to make it a fixed solution?

  26. anonymous
    • 5 years ago
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    yups..u got it!

  27. anonymous
    • 5 years ago
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    omg thank you so much for being so patient with me :) medal

  28. anonymous
    • 5 years ago
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    thnx..ask if u hav any probs..good luck!

  29. anonymous
    • 5 years ago
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    ok thx! do you apply the same process when talking about vertices and trying to form triangles and diagonal lines?

  30. anonymous
    • 5 years ago
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    no...i dont think so..:) its only for letters btw...for that kind of problems i know how to do them..ask me if u want!

  31. anonymous
    • 5 years ago
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    sorry for imposing :( the question was: A regular polygon was 10 vertices. use combinatorics to find: triangles that could be formed using the vertices diagonals that could be drawn

  32. anonymous
    • 5 years ago
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    a regualr polygon with 10 vertices means no 3 points are in a line..i.e. they are not collinear. So we have triangle from any 3 points from among the 10 points. So that means we have to find the number of ways we can select 3 points from 10 points.i.e. 10C3.This gives the number of triangles.

  33. anonymous
    • 5 years ago
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    For diagonals, we man by diagonals any line inside the polygon. So we cant take sides as diagonals. We get a line by joining any 2 points. But evry line is not a diagonal..see that the sides are also lines obtained by joining 2 points...so we find the (total lines)-(total sides) i.e. 10C2-10.which gives the no. of diagonals..hope u understood:)

  34. anonymous
    • 5 years ago
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    got it smoochumkisses?

  35. anonymous
    • 5 years ago
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    yes it makes sense when you put it conceptually as the points aren't colinear so then would the diagonal lines question be 10C2 / 2! to account for the diagonal line coming from two directions?

  36. anonymous
    • 5 years ago
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    no..it will be 10C2-10. Not 10C2/2! since in combinations we take one arrangement of a particular pair...u could have said 10P2/2!-10 though..its the same as my answer..

  37. anonymous
    • 5 years ago
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    -10? why?

  38. anonymous
    • 5 years ago
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    i told u in my post. we must neglect the sides of the polygon. Sides are not diagonals..:)

  39. anonymous
    • 5 years ago
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    oh okay, so you don't divide by 2! because it doesn't matter the order just as long as the two points are selected still but you subtract 10 because the sides can't be diagonals

  40. anonymous
    • 5 years ago
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    you got that!! congr8s! :P

  41. anonymous
    • 5 years ago
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    you make it easier to understand XD thx u time to do some binomial expansion problems now hehe thanks soo much again

  42. anonymous
    • 5 years ago
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    ask me if u have any problems for expansions:) happy to help:)

  43. anonymous
    • 5 years ago
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    what math are you in at school btw?? this is all pre-calc but im sure you could tell XD

  44. anonymous
    • 5 years ago
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    i have passed high school yesterday...:) got 100/100 in maths..

  45. anonymous
    • 5 years ago
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    oh grats on passing :] omg wicked XD 100

  46. anonymous
    • 5 years ago
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    thnx..what math r u in and in which grade?

  47. anonymous
    • 5 years ago
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    pre-calc freshman

  48. anonymous
    • 5 years ago
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    k.gud! best of luck..

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