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 Three from Nine (Posted on 2006-04-25)
Nine marbles numbered 1 to 9 are placed in a barrel and three are drawn out, without replacement. Determine -:

1. The probability that the three digit number formed from the marbles in the order drawn is divisible by (a) Five (b) Seven (c) Nine.
2. The probability that a three digit number can be formed by rearranging the marbles drawn, that is divisible by (a) Five (b) Seven (c) Nine.

 No Solution Yet Submitted by Vernon Lewis Rating: 3.5000 (4 votes)

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 re: Partial Solution - ? with no answer of my own | Comment 5 of 12 |
(In reply to Partial Solution by tomarken)

If the first two numbers total to 9 (i.e. 3 and 6) a zero marble is not available to keep the probability as 1/7.

If the first two numbers are 2 and 5, then another 2 is needed but not available.

I'm concerned that this may be a brute force type of problem.

Edited on April 25, 2006, 1:24 pm
 Posted by Leming on 2006-04-25 13:22:34

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