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anonymous

  • one year ago

HELP PLEASE:) FAN&MEDAL&TESTIMONIAL:):) A committee of 5 people is to be chosen from a group of 8 women and 10 men. The committee must feature 3 women and two men? The committee must have more women than men?

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  1. anonymous
    • one year ago
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    Without restrictions all the possibilities are accounted for by 18P5

  2. anonymous
    • one year ago
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    but with the above restrictions.....:(

  3. ganeshie8
    • one year ago
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    Wrong, does the order in which you pick people matter here ?

  4. anonymous
    • one year ago
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    I don't think so.

  5. ganeshie8
    • one year ago
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    The order in which you select the people doesn't matter here because, end of the day, all you bother about is "who" are in the committee. Not in which order you picked them.

  6. ganeshie8
    • one year ago
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    So this better be a combination problem, yes ?

  7. anonymous
    • one year ago
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    Yes.

  8. ganeshie8
    • one year ago
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    Without restrictions all the possibilities are accounted for by 18\(\color{red}{C}\)5

  9. anonymous
    • one year ago
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    What's the difference between P and C? I am sorry I am not very advanced

  10. ganeshie8
    • one year ago
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    Maybe consider a quick example : Suppose you went to juice shop and ask for a salad made up of "bananas, apple, grapes"

  11. ganeshie8
    • one year ago
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    does the order in which you ask matter here ?

  12. anonymous
    • one year ago
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    No. You can go about picking any

  13. anonymous
    • one year ago
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    But what difference does it make if permutation changed to combination?

  14. anonymous
    • one year ago
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    Lack of order?

  15. ganeshie8
    • one year ago
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    `give me a salad made up of "bananas, apple, grapes"` is no different from `give me a salad made up of "apple, bananas, grapes"` order doesn't matter here, so this is a combination problem.

  16. anonymous
    • one year ago
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    ok I get it now.

  17. ganeshie8
    • one year ago
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    consider another quick example : you want to choose four digits to set a password to your laptop hard disk : "8 7 3 9" does the order matter here ?

  18. ganeshie8
    • one year ago
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    sorry to persist, but i think this is really important..

  19. anonymous
    • one year ago
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    No

  20. ganeshie8
    • one year ago
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    Are you saying "8 7 3 9" is no different from "7 3 8 9" ?

  21. ganeshie8
    • one year ago
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    does your computer accept/treat both passwords as same ?

  22. anonymous
    • one year ago
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    No

  23. anonymous
    • one year ago
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    The computer only accepts a given password you made at the boot up

  24. anonymous
    • one year ago
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    So the order would have to remain the same.

  25. ganeshie8
    • one year ago
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    Exactly, since the "order" in which you enter the digits matter here, this is clearly a permutation problem.

  26. ganeshie8
    • one year ago
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    That is how you differentiate between a permutation and combination problem

  27. anonymous
    • one year ago
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    Oh ok. I get it now.

  28. anonymous
    • one year ago
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    Ordinal or nominal

  29. ganeshie8
    • one year ago
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    ``` A committee of 5 people is to be chosen from a group of 8 women and 10 men. The committee must feature 3 women and two men? ``` How many ways can you choose 3 women from 8 women ?

  30. anonymous
    • one year ago
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    24

  31. ganeshie8
    • one year ago
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    good guess, but no.

  32. anonymous
    • one year ago
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    Either this is too hard or I am too stupid:(

  33. anonymous
    • one year ago
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    2.7 something

  34. anonymous
    • one year ago
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    8/3=2.66

  35. ganeshie8
    • one year ago
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    Neither, you just don't happen to have enough practice. You can think of this exactly same as the permutation problem. 3 women can be chosen from 8 women in 8C3 ways

  36. ganeshie8
    • one year ago
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    where \(nCr\) is defined as \[nCr = \dfrac{n!}{r!*(n-r)!}\]

  37. anonymous
    • one year ago
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    8C3=number of possibilities where 3 women are picked from 8 women.

  38. ganeshie8
    • one year ago
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    \[8C3 = \dfrac{8!}{3!*(8-3)!} = \dfrac{8!}{3!*5!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1*5\times 4\times 3\times 2\times 1}\]

  39. ganeshie8
    • one year ago
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    Yes, see if you can simplify above

  40. anonymous
    • one year ago
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    (8*7*6*5)/3*2*1=336 1680/6=

  41. anonymous
    • one year ago
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    Because most numbers cancel each other out.

  42. anonymous
    • one year ago
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    8*7*6/3*2*1

  43. ganeshie8
    • one year ago
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    Yes simplifies further

  44. anonymous
    • one year ago
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    330/6=55

  45. ganeshie8
    • one year ago
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    8*7*6/3*2*1 8*7 56 right ?

  46. anonymous
    • one year ago
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    yes.

  47. anonymous
    • one year ago
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    so 56 ways in which 3 women can be picked from 8 women.

  48. ganeshie8
    • one year ago
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    Yes next find how many ways 2 men can be picked from the available 10 men

  49. anonymous
    • one year ago
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    2C10 so

  50. ganeshie8
    • one year ago
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    10C2 ok

  51. ganeshie8
    • one year ago
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    work it out

  52. anonymous
    • one year ago
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    10!/2!(10-2)!

  53. anonymous
    • one year ago
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    10!/2!*8!

  54. anonymous
    • one year ago
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    10*9*8*7*6*5*4*3*2*1/2*1*8*7*6*5*4*3*2*1*

  55. anonymous
    • one year ago
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    I simplify that

  56. ganeshie8
    • one year ago
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    yes do it

  57. anonymous
    • one year ago
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    90/2=45

  58. anonymous
    • one year ago
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    45 ways in which 2 men can be picked from 10 men

  59. anonymous
    • one year ago
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    56*45 is my guess.

  60. ganeshie8
    • one year ago
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    Excellent! so a committee with 3 women and two men can be formed in : \[8C3*10C2 = 56*45~\text{ways}\]

  61. anonymous
    • one year ago
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    I got it:)

  62. anonymous
    • one year ago
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    Do I use the same for the last one?

  63. anonymous
    • one year ago
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    where there are more men than women.

  64. anonymous
    • one year ago
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    there are 3 cases where men can be more than women 3 men 2 women 4 men 1 woman 5 men 0 women

  65. anonymous
    • one year ago
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    Sorry reverse of that

  66. anonymous
    • one year ago
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    more women than men. So do I multiply our previous answer by 3?

  67. ganeshie8
    • one year ago
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    sry was away.. still here @Robert136

  68. ganeshie8
    • one year ago
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    ``` there are 3 cases where men can be more than women 3 women 2 men 4 women 1 man 5 women 0 man ``` Yes find the number of ways for each case, then add up.

  69. ganeshie8
    • one year ago
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    you should get \[8C3*10C2 + 8C4*10C1+5C1*10C0\]

  70. ganeshie8
    • one year ago
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    wolfram says that evaluates to 3225 http://www.wolframalpha.com/input/?i=%288+choose+3%29*%2810+choose+2%29+%2B+%288+choose+4%29*%2810+choose+1%29%2B%285+choose+1%29*%2810+choose+0%29

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