A member of a book club wishes to purchase two books from a selection of eight books recommended for a certain month. In how many ways can she choose them?
${ }_n C_r=\frac{n!}{r!(n-r)!}$
Answer (1)
Recognize the problem as a combination since the order of book selection doesn't matter.
Apply the combination formula: n C r = r ! ( n − r )! n ! with n = 8 and r = 2 .
Calculate 8 C 2 = 2 ! 6 ! 8 ! = 2 × 1 8 × 7 .
Simplify to find the number of ways: 28 .
Explanation
Understand the problem We are given that a member of a book club wants to purchase two books from a selection of eight books. We need to find the number of ways she can choose them. Since the order in which the books are chosen does not matter, this is a combination problem.
State the combination formula The combination formula is given by: n C r = r ! ( n − r )! n ! where n is the total number of items, and r is the number of items to choose. In this case, we have n = 8 (total number of books) and r = 2 (number of books to choose).
Substitute the values Substituting the values into the formula, we get: 8 C 2 = 2 ! ( 8 − 2 )! 8 ! = 2 ! 6 ! 8 ! Now, let's calculate the factorials.
Calculate the factorials We have: 8 ! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 2 ! = 2 × 1 = 2 6 ! = 6 × 5 × 4 × 3 × 2 × 1 Substituting these values back into the combination formula: 8 C 2 = 2 × 6 ! 8 × 7 × 6 ! We can cancel out the 6 ! from the numerator and the denominator.
Simplify and calculate the final result 8 C 2 = 2 8 × 7 = 2 56 = 28 Therefore, there are 28 ways to choose two books from a selection of eight books.
Examples
In a classroom of 25 students, a teacher wants to select a group of 3 students to help with a project. This is a combination problem because the order in which the students are selected does not matter. The number of ways to choose 3 students from 25 is given by 25 C 3 = 3 ! ( 25 − 3 )! 25 ! = 3 ! 22 ! 25 ! = 3 × 2 × 1 25 × 24 × 23 = 25 × 4 × 23 = 2300 . So, there are 2300 different groups of 3 students that can be selected.
