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Topic: Permutation Combination 2 (Read 398 times) 

navdeep1771
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Permutation Combination 2
« on: Jun 18^{th}, 2018, 9:57pm » 
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There are four basketball players A, B, C, D. Initially, the ball is with A. The ball is always passed from one person to a different person. In how many ways can the ball come back to A after seven passes? (For example A > C > B > D > A > C > D > A ) is one way in which ball can come back to A.

« Last Edit: Jun 18^{th}, 2018, 10:00pm by navdeep1771 » 
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rmsgrey
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Re: Permutation Combination 2
« Reply #1 on: Jun 19^{th}, 2018, 6:18am » 
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Answer: 546 Solution: hidden:  Letting A(n) be the number of ways for A to have the ball after n passes, and B(n) ( =C(n)=D(n) ) be the number of ways for B (or C or D) to have it after n passes gives the recurrence relation: A(0) = 1 B(0) = 0 A(n) = B(n1)+C(n1)+D(n1) = 3B(n1) B(n) = A(n1)+C(n1)+D(n1) = A(n1) + 2B(n1) Iterating gives: n: A(n), B(n) 0: 1, 0 1: 0, 1 2: 3, 2 3: 6, 7 4: 21, 20 5: 60, 61 6: 183, 182 7: 546, 547 


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towr
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Re: Permutation Combination 2
« Reply #2 on: Jun 20^{th}, 2018, 10:53am » 
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One might simplify A(n) further to A(n) = 2A(n1) + 3A(n2) = [3*(1)^n + 3^n)]/4


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Grimbal
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Re: Permutation Combination 2
« Reply #3 on: Jun 21^{st}, 2018, 2:17pm » 
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If a_{0} counts the ways for A to have the ball and a_{1} the ways not to have the ball, then you start with a={1,0}. If A has the ball, after a pass, A has 0 ways to have the ball and 3 ways not to have it. If A does not have the ball, after a pass, A has 1 way to have the ball and 2 ways not to have it. So a pass multiplies the vector 'a' by {{0,1},{3,2}}. After 7 passes it is {{0,1},{3,2}}^7 = {{546,547},{1641,1640}}. See http://www.wolframalpha.com/input/?i={{0,1},{3,2}}^7 So the number of ways is: {1,0} x {{0,1},{3,2}}^7 x {1,0} = 546. And the formula works for other values of the constant 7. PS: The same page gives the eigenvalues of the matrix as 3 and 1. So you know the general formula for A(n) is a combination of 3^n and 1^n. As towr found out.

« Last Edit: Jun 21^{st}, 2018, 2:37pm by Grimbal » 
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towr
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Re: Permutation Combination 2
« Reply #4 on: Jun 21^{st}, 2018, 11:04pm » 
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We can generalize this problem to P players, and players having to wait T turns before the ball can return to them, with the sequences of passes ending after N turns. (So we've dealt with P=4,T=1,N=7) Is there a simple generic solution for this? (Taking P into account seems simple enough, but maybe T might complicate things)

« Last Edit: Jun 21^{st}, 2018, 11:04pm by towr » 
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Grimbal
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Re: Permutation Combination 2
« Reply #5 on: Jun 22^{nd}, 2018, 4:39am » 
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One difficulty is that for instance the first time A throws the ball he can throw it to all P1 players. But the second time, there are T players (A inclusive) who can not receive the ball. So the first T1 throws must be treated specially, the players have more choices than the (PT) choices they have eventually.


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towr
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Re: Permutation Combination 2
« Reply #6 on: Jun 22^{nd}, 2018, 5:27am » 
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Don't the first T1 turns just give a multiplier of (P1)!/(T1)!, and from that point on you can treat it generically as (PT) choices onward?


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Grimbal
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Re: Permutation Combination 2
« Reply #7 on: Jun 23^{rd}, 2018, 3:46am » 
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hidden:  Right you are! The first T turns A will not receive the ball anyway. The vector must now describe how long A did not have the ball, from 0 to T turns. It has T+! dimensions. After T turns, the vector has a 1 at position T. a = {0,...,1}. From there on, the vector is multiplied by the following matrix: (shown for T=3). (  0  0  0  1  )  (  PT  0  0  0  )  (  0  PT  0  0  )  (  0  0  PT  PT1  )  We can assume P>T because else the process has to stop before A gets the ball back. Since the information on the starting position gets diluted fast, the number of ways for A to get the ball in N turns should approach (the total number of paths in N turns)/(#players) That is ways ~= (P1)! / (PT1)! * (PT)^(NT) / P 

« Last Edit: Jun 23^{rd}, 2018, 3:47am by Grimbal » 
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