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P&C
Asked by | 16 Feb, 2010, 11:14: PM Expert Answer

(c) 1630

We can pick the numbers, one at a time, two.. upto six.

So total number of positive intergers possible 6P1 + 6P2 + 6P3 + 6P4 + 6P5 + 6P6 = 1956

Total number of positive integers starting with zero =   5P1 + 5P2 + 5P3 + 5P4 + 5P5 = 325

Total number of positive integers = 1956 - 325 - 1 = 1630,

(Note that we have subtracted 1, since we don't want 0 to be counted as a positive integer)

Regards,

Team,

TopperLearning.

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