One in a zillion

Factual Questions
1

I’ve recently stumbled across one of those lottery Web sites. In this case, all you have to do to win the Big Prize ($3,000,000 – $75,000 per year for 40 years, with no interest), is select 6 numbers out of … 94. In the Rules section, they give you your odds: 814,216,767 to one. Which probably makes you several thousands times more likely to be struck by lightning twice in the same spot holding a steel rod on the top of your head.

But being the kind souls that they are, they add that picking order does not matter. In other words, if you chose 2, 18, 43, 7, 19, 94 and these were the winning numbers drawn albeit in a different order, you’d win, no matter what.

Here’s my question to the math/probability wiz kids in the Doper Community: What would the odds be if the picking order DID matter? If the balls (?) had to come up in the exact same sequence. Is there a way of calculating this with any degree of certainty?

2

There are 6 * 5 * 4 * 3 * 2 = 720 (usually writen 6! or 6 factorial) ways of selecting 6 balls. Therefore if order is important multiply the result by 720. i.e. 586236072240

BTW the original figure was calculated by 94!/ ((94 -6)! * 6!)

3

Thanks ticker.

(Keeping my fingers crossed for the Big One…:D:D)