1)
A lottery is set up in which players pick six numbers from the set 1, 2, 3, ... , 39, 40. How many different ways are there to play this lottery? (In this game the order in which the numbers are picked does NOT matter.)
2)
A club of 15 students will form a subcommittee consisting of three students. How many different such subcommittees can be formed?
1)40C6=40!/(6!x34!)=3,838,380
2)15C3=15!/(3!x12!)=455
1) To calculate the number of different ways to play the lottery, we need to calculate the number of combinations. Since the order of the numbers does not matter, we can use the combination formula.
The formula for combinations is:
C(n, r) = n! / (r!(n - r)!)
Where n is the total number of options and r is the number of choices we need to make.
In this case, there are 40 options to choose from (numbers 1 to 40), and we need to choose 6 numbers. Plugging these values into the combination formula:
C(40, 6) = 40! / (6!(40 - 6)!)
Simplifying:
C(40, 6) = 40! / (6! * 34!)
Using factorials can be computationally expensive, so we can simplify further by dividing the numerator and denominator by their respective factorials:
C(40, 6) = (40 * 39 * 38 * 37 * 36 * 35) / (6 * 5 * 4 * 3 * 2 * 1)
Cancelling out common factors:
C(40, 6) = (8 * 13 * 19 * 37 * 7) / 1
Calculating the result:
C(40, 6) = 3,838,380
Therefore, there are 3,838,380 different ways to play this lottery.
2) To calculate the number of different subcommittees that can be formed, we can again use the combination formula:
C(n, r) = n! / (r!(n - r)!)
In this case, there are 15 students to choose from, and we need to choose 3 students. Plugging these values into the combination formula:
C(15, 3) = 15! / (3!(15 - 3)!)
Simplifying:
C(15, 3) = 15! / (3! * 12!)
Cancelling out common factors:
C(15, 3) = (13 * 14 * 15) / (3 * 2 * 1)
Calculating the result:
C(15, 3) = 455
Therefore, there are 455 different subcommittees that can be formed.