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April 16, 2014

April 16, 2014

Posted by **BreAnna** on Saturday, April 16, 2011 at 11:21pm.

- math help pls! -
**MathMate**, Sunday, April 17, 2011 at 8:28amWhen the order does not matter, the number ways of choosing the lottery can be calculated using combinations, or C(n,r), read as "n chose r". C(n,r) is defined as n!/(r!(n-r)!), similar to the evaluation problem of your other post.

So in the case of the lottery problem, n=40, r=6, and order does not matter. So "40 choose 6" gives the answer.

Evaluate "40 choose 6" the same way as the other problem:

40!/(6!(40-6)!)=....

Read the following link for more detailed information:

http://en.wikipedia.org/wiki/Combination

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