Roughly 195 million. Oh dear! At least that's the jackpot odds. When you hear that you can win 50 or 100 million dollars if you play $1, your expected value (prize * expected probability of winning) is still less than $1.
I'm assuming they're three unique numbers. Thus, the first can be any of three, the second either of the remaining two, and the last is the last one left. Thus: combinations = 3 * 2 * 1 = 6 Or, more...
16 if you include 0 starting12 if notThat is not correct.0123 is a 4-digit number = (n)therefore n! = 4! = 1*2*3*4 = 24 (if 0 is included as I assume that is the question)123 is a 3-digit number =...
In general, the number of combinations of n things taken r at a time is nCr = n!/[(n - r)!r!] Thus, we have: 10C8 = 10!/[(10 - 8)!8!] = 10!/(2!!8!) = (10 x 9 x 8!)/(8! x 2 x 1) = (10 x 9)/2 = 5 x 9 =...