Since the order does not matter, we are "choosing" 3 out of the 9,
that would be C(9,3) = 9!/(3!6!) = 84
HELP....
In a certain lottery, 3 numbers between 1 and 9 inclusive are drawn. These are the winning numbers. How many different selections are possible? Assume the order in which the numbers are drawn is not important.
Answer choice: a.) 6 b) 84 c) 504 d) 729
3 answers
9!/(3!*6!) = 7*8*9/1*2*3 = ?
4#OUT OF 9#=ORDER?
4#OUT OF9#=ANY ORDER?
EXM:1234?
4#OUT OF9#=ANY ORDER?
EXM:1234?
1 answer