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January 29, 2015

January 29, 2015

Posted by **anonymous** on Monday, October 8, 2007 at 10:00am.

- math -
**drwls**, Monday, October 8, 2007 at 10:36amEach permutaion of five numbers, following the first, has 120 possibilities. Therefore there are 360 numbers begining with 1, 2, or 3, and you are looking for the 49th number beginning with 4.

4! = 24 numbers will begin with each two-number combination. 361 through 384 will begin with 41. 385 through 408 will begin with 42. Therefore your number will be the first (smallest) number begining with 43.

What would that be?

431256

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