Permutation question

• Diehard_89

  9 Jan 07, 20:28

  how many ways can 9 balls of 4red 4white and 1black be arranged in a line such that no red ball is next to the black
  
  any one?
• bryanseb

  9 Jan 07, 20:31

  Originally posted by Diehard_89:

  how many ways can 9 balls of 4red 4white and 1black be arranged in a line such that no red ball is next to the black
  
  any one?

  Hmmm.
  
  R = Red
  W = White
  B = Black
  
  R, W, R, W, R, W, R, W, R, W, B
• Unidentified

  9 Jan 07, 20:44

  no red ball is next to the black ?
  
  R , R , W , W , B , W , W , R , R
• Diehard_89

  9 Jan 07, 20:45

  yeap I know
  
  but how many ways are there?
• snop_coolz

  9 Jan 07, 20:48

  B W R W R W R W R
  
  W B W R W R W R R
  
  W W W B W R R R R
  
  alot wat...
  
  9! ... donno correct or not..
• LyonK

  9 Jan 07, 21:17

  
• mir4cle

  9 Jan 07, 21:36

  LOlx..dots..secondary school qn..cant realli remember..sorry..i noe is dono wad ! den multiply by donno wad ! ..
• snop_coolz

  9 Jan 07, 21:38

  donno..u try 9!..or u have to slowly draw lor..
  
  
• ndmmxiaomayi

  9 Jan 07, 21:40

  Deleted.
• c0mplex

  9 Jan 07, 21:43

  Originally posted by Diehard_89:

  how many ways can 9 balls of 4red 4white and 1black be arranged in a line such that no red ball is next to the black
  
  any one?

  hmm, first, find total number of ways you can arrange all the balls(without the restrictions). you should know how to this if you ask this question. let it be x
  
  next, find the number of ways a red ball will be next to a black ball. this can be done by making the red and black ball a new 'ball'. this means that you will have 8 balls now, 3 red, 4 white and 1 red/black ball. you can then find the number of ways you can arrange those balls. let it be y. then, within the new 'ball', you can have to ways to arrange it(RB or BR). so, the total number of ways a red ball will be next to a black ball is 2y
  
  so, total number of ways such that no red ball is next to the black = x - 2y
• hegu

  9 Jan 07, 21:47

  sry....deleted for revision of answer
• -purpose

  9 Jan 07, 21:47

  erms...9!-(2!x8!) = 282240??
  i think..lol
• praetor81

  9 Jan 07, 21:50

  PUT UR OWN BALLS IN BETWEEN
• peeehead

  9 Jan 07, 21:58

  Originally posted by Diehard_89:

  how many ways can 9 balls of 4red 4white and 1black be arranged in a line such that no red ball is next to the black
  
  any one?

  arrangement: black, white,red,white,red,white,red,white,red.
  
  the way I do it: 4 x 3 x 2 x 1 PLUS 4 x 3 x 2 x 1 OR 4! plus 4!
  
  I THINK
  
  so answer is 48. (?)
• angelfairy

  9 Jan 07, 22:00

  350
• angelfairy

  9 Jan 07, 22:02

  Originally posted by peeehead:

  arrangement: black, white,red,white,red,white,red,white,red.
  
  the way I do it: 4 x 3 x 2 x 1 PLUS 4 x 3 x 2 x 1 OR 4! plus 4!
  
  I THINK
  
  so answer is 48. (?)

  what about white black white
• peeehead

  9 Jan 07, 22:04

  Originally posted by angelfairy:

  what about white black white

  what you mean white black white?
  only one black. then red cannot be together with black.
  if there's another possibility, then it's the second case already.
  so must plus together with thi sone
• hegu

  9 Jan 07, 22:06

  7!/(2! x 4!) + 7!/(3! x 4!) x 2 = 175
  
  70% confident....
  
  TS do you have the final answer? so at least we know where we are correct or not
• peeehead

  9 Jan 07, 22:07

  i know already.
  no red ball next to the black.
  
  9! - (red ball next to the black)
  
  
• c0mplex

  9 Jan 07, 22:11

  Originally posted by peeehead:

  i know already.
  no red ball next to the black.
  
  9! - (red ball next to the black)
  
  

  it's not 9!, remember that the same coloured balls are identical...
• hegu

  9 Jan 07, 22:13

  Originally posted by peeehead:

  i know already.
  no red ball next to the black.
  
  9! - (red ball next to the black)
  
  

  calculating the no. of combi for (red ball next to black) is harder than you think
  
  cos too many overlapping cases
• c0mplex

  9 Jan 07, 22:14

  Originally posted by hegu:

  7!/(2! x 4!) + 7!/(3! x 4!) x 2 = 175
  
  70% confident....
  
  TS do you have the final answer? so at least we know where we are correct or not

  answer should be 70... i'm 90% sure...
• angelfairy

  9 Jan 07, 22:14

  Originally posted by peeehead:

  what you mean white black white?
  only one black. then red cannot be together with black.
  if there's another possibility, then it's the second case already.
  so must plus together with thi sone

  meaning you only considered when the black was at the corner
• Diehard_89

  9 Jan 07, 22:19

  hmm thanks guys
  
  I don't have the final answer but I get different answer
  
  what I do was
  
  I find the one which has no restriction = 630
  
  and I find the no. of ways whereby the red and blacks are together, meaning at one point one of the reds is in contact with the black = 25
  
  so in the end I get 605
• hegu

  9 Jan 07, 22:20

  Originally posted by hegu:

  7!/(2! x 4!) + 7!/(3! x 4!) x 2 = 175
  
  70% confident....
  
  TS do you have the final answer? so at least we know where we are correct or not

  case 1:
  
  (W B W) (6 other balls) => 7 groups to permutate altogether. Treat (W B W) as one group
  
  combi = 7!/(2! x 4!) = 105 (need to divide becos same coloured balls are identical)
  
  case 2a:
  
  (B W) (7 other balls) => 7 groups to permutate altogether. (B W) group is fixed on the LHS
  
  case 2a:
  
  (7 other balls) (W B) => 7 groups to permutate altogether. (W B) group is fixed on the RHS
  
  summing up case 2a and 2b, combi = 7!/(3! x 4!) x 2 = 70
  
  add these 3 cases you will get 175.
  
  correct me if i am wrong....or i missed out any cases or overlaps