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About 'choose' function

On 08-Nov-04 Ted Harding wrote:
I just did this calculation using 'bc' (which works to arbitrary
decimal precision), getting (after rounding 59999999999.99...996,
1744 9's in all, to 60000000000):

16131211663389665874291103209859887097643782097733433535176164529013\
93347758720756812586746724689691665340326913939172715147819303293449\
54257506586926613288215108141984988883561331127046497391143349652601\
81932350377398578090978006727056282867666693363201127636940304448887\
82022049197129032260000000000

so the leading digits are 1613121..., agreeing with 'my.choose3'
as above, rather than with 'choose'!

Ted.


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Date: 08-Nov-04                                       Time: 15:11:37
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