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Sum (Pair Product) = Sum (Triplet Product) (Posted on 2009-11-08) Difficulty: 2 of 5
Determine the probability that for a positive integer N chosen at random between 1000 and 9999 inclusively, the sum of the products of pairs of digits in N is equal to the sum of products of triplets of its digits.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Question questions? | Comment 1 of 3
Perhaps I'm missing something but I'm having difficulty understanding exactly what is meant by sum of product of pairs and triples.  Is the the sum of the product of every possible way of choose 2 or 3 digits? If so is this only the sum of the unique pairs and triplets?  For instance with 1233 would it be 1*1+1*2+1*3+1*3+2*1+2*2+2*3+2*3+3*1+3*2+3*3+3*3
+3*1+3*2+3*3+3*3 or would we discard duplicates and if so are we considering the pairs as ordered (is 1*2 the same as 2*1).
  Posted by Daniel on 2009-11-08 14:51:02
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