Solve by the method of choice
A book club offers a choice of 8 books from a list of 40. in how many ways can a member make a selection?
I will assume that the books are different
Number of ways to choose 8 books from 40
= C(8,4) = 8!/(4!4!) = 70
(order of choosing the books should not matter)
To solve this problem, we can use the concept of combinations. A combination is a way to select a specified number of items from a larger set without regard to the order of the items. In this case, we want to find the number of ways a member can select a certain number of books from a list.
We can use the formula for combinations, which is:
C(n, r) = n! / (r!(n-r)!)
where:
- n is the total number of items in the set (40 books in this case),
- r is the number of items we want to select (8 books in this case), and
- ! denotes the factorial of a number, which is the product of all positive integers less than or equal to that number.
Plugging in the values into the formula, we get:
C(40, 8) = 40! / (8!(40-8)!)
Simplifying the equation, we have:
C(40, 8) = 40! / (8! * 32!)
Calculating the factorials:
40! = 40 * 39 * 38 * ... * 3 * 2 * 1
8! = 8 * 7 * 6 * ... * 3 * 2 * 1
32! = 32 * 31 * 30 * ... * 3 * 2 * 1
Now we can substitute these values into our equation:
C(40, 8) = (40 * 39 * 38 * ... * 3 * 2 * 1) / [(8 * 7 * 6 * ... * 3 * 2 * 1) * (32 * 31 * 30 * ... * 3 * 2 * 1)]
After simplifying the equation further by canceling out common terms, we should be left with the final answer, which represents the number of ways a member can make a selection from the book list.