in a certain lottery, 3 numbers between 1 and 9 inclusive are drawn. These are the winning numbers. How many different selections are possible?
Does the order of draw matter? (Usually not)
Can a number be drawn more than once? (Usually not)
In the "usual" lottery situation, the answer for this case is
9!/(6!*3!) = 9*8*7/(1*2*3) = 84
To calculate the number of different selections possible in this lottery, we need to determine the number of combinations of 3 numbers that can be drawn from a set of 9.
To find the number of combinations, we can use the formula for combinations, which is given by:
C(n, r) = n! / (r! * (n-r)!)
Where n is the total number of items in the set and r is the number of items being chosen at a time.
In this case, we have a set of 9 numbers (1 to 9) and we need to choose 3 numbers at a time. Therefore, the formula becomes:
C(9, 3) = 9! / (3! * (9-3)!)
Simplifying the equation:
C(9, 3) = 9! / (3! * 6!)
The exclamation mark (!) represents the factorial function, which means multiplying a number by all positive integers less than it down to 1.
Calculating the factorials:
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
3! = 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
Substituting the factorials into the equation:
C(9, 3) = (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (6 * 5 * 4 * 3 * 2 * 1))
Cancelling out common terms in the numerator and denominator:
C(9, 3) = (9 * 8 * 7) / (3 * 2 * 1)
Calculating the result:
C(9, 3) = 84
Therefore, there are 84 different selections possible in this lottery.