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
Since the order does not matter, we are "choosing" 3 out of the 9,
that would be C(9,3) = 9!/(3!6!) = 84
9!/(3!*6!) = 7*8*9/1*2*3 = ?
4#OUT OF 9#=ORDER?
4#OUT OF9#=ANY ORDER?
EXM:1234?
To find the number of different selections possible in the lottery, we can use the concept of combinations.
In this scenario, we need to select 3 numbers out of 9 numbers (from 1 to 9 inclusive), without considering the order in which the numbers are drawn.
The formula for combinations is given by:
C(n, r) = n! / (r! * (n - r)!)
Where:
- n is the total number of items to choose from
- r is the number of items to be chosen
In this case, n = 9 (the range of numbers to choose from) and r = 3 (the number of numbers to be chosen).
Using the formula, we can calculate the number of selections:
C(9, 3) = 9! / (3! * (9 - 3)!)
= 9! / (3! * 6!)
= (9 * 8 * 7 * 6!) / (3! * 6!)
= (9 * 8 * 7) / 3!
= 504 / 6
= 84
Therefore, there are 84 different selections possible for the winning numbers.
So, the correct answer is b) 84.