Ex: You have five books and are going to select three are to read. How many different combinations are there, ignoring the order in which they are selected?

To solve this problem, we can use the combination formula, which is given by:<br /><br />\[ C(n, k) = \frac{n!}{k!(n-k)!} \]<br /><br />where \( n \) is the total number of items, and \( k \) is the number of items to choose.<br /><br />In this case, \( n = 5 \) (the total number of books) and \( k = 3 \) (the number of books to select).<br /><br />Plugging these values into the formula, we get:<br /><br />\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} \]<br /><br />Calculating the factorials, we have:<br /><br />\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]<br />\[ 3! = 3 \times 2 \times 1 = 6 \]<br />\[ 2! = 2 \times 1 = 2 \]<br /><br />Substituting these values back into the formula, we get:<br /><br />\[ C(5, 3) = \frac{120}{6 \times 2} = \frac{120}{12} = 10 \]<br /><br />Therefore, there are 10 different combinations of 3 books that can be selected from 5 books, ignoring the order in which they are selected.