Let k be a positive integer. Suppose that the integers 1, 2, 3, ...3k, 3k + 1 are written down in random order.
What is the probability that at no time during this process, the sum of the integers that have been written up to that time is divisible by 3?
Source: Putnam competition
(In reply to
re(3): Solution by Charlie)
I found it:
there are (k!*(k+1)!*(3k)!)/((2k)!*k!) = (k+1)!*(3k)!/((2k)!) ways to create a desires sequence out of all (3k+1)! possible sequences.
The first (k!*(k+1)!*(3k)!)/((2k)!*k!) uses the combination formula for C(3k,k) for the positioning of the multiples of 3, rather than using permutations. That means 1346572 and 1643572 count only once in the numerator, but twice in the (3k+1)! denominator of overall permutations of the 7 (in this instance) numbers. The multiples of 3 are not identical and their order counts just as much in the numerator as in the denominator.
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Posted by Charlie
on 2016-03-17 23:21:55 |
re(4): Solution
