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January 26, 2015

January 26, 2015

Posted by **JOHN** on Wednesday, June 18, 2014 at 6:12pm.

- ALGERBRA -
**MathMate**, Wednesday, June 18, 2014 at 6:43pmNumber of possible combinations of 5 (distinct) numbers out of 30

= 30C5

Probability of winning if n different tickets are bought

= n/(30C5)

where 30C5

=30!/((30-5)!5!)

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