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BlackBerry Limited Message Board

  • joelew111 joelew111 Mar 7, 2013 2:10 PM Flag

    Thor relleases another clue to device sales numbers...I don't get it.

    A boy has four white Z10's and eight black Z10's. He arranges his twelve Z10's randomly, in a ring. What is the probability that no black or white Z10's are ajacent?

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    • The chances are 100% in favour of two black Z10s being together in two places.

      Sentiment: Hold

    • Start smashing your head against a brick wall that riddle is just unsolvable.

      • 1 Reply to midnightrambler108
      • This problem is a counting exercise. We will count the number of distinct Z10 arrangements such that no two white Z10’s are adjacent, and then divide this by the total number of distinct Z10 arrangements to obtain the required probability.

        To simplify the counting process, select any black Z10, and consider the remaining 11 Z10’s, arranged in a line. The proportion of such arrangements for which no two white Z10’s are adjacent will be the same as for the original 12 z10’s, arranged in a ring.

        The number of ways of choosing k outcomes out of n possibilities, ignoring order, nCk, is equal to n! / [k! (n − k)!].
        So the total number of ways of arranging 4 white Z10’s out of 11 is 11C4 = 330.

        To count the number of arrangements such that no two Z10s are adjacent, observe that there must be at least one black Z10 between each two would-be adjacent white Z10:

        [red][blue][red][blue][red][blue][red]

        Having fixed the positions of three black Z10, we have four black Z10 left to play with. We can think of the four white Z10’s as dividing lines, around which the remaining four black Z10’s must be slotted. Given four dividing lines and four Z10’s, we have 8C4 = 70 distinct combinations.

        Therefore the probability that no two white Z10’s are adjacent is 70/330 = 7/33.

 
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