\( 1 \leftarrow \quad \) A certain lottery has 30 numbers. In how many different ways can 4 of the numbers be selected? (Assume that order of selection is not important.)

To determine the number of ways to select 4 numbers from a set of 30 without regard to the order, we use the combination formula, which is given by: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \( n \) is the total number of items to choose from (in this case, 30), \( r \) is the number of items to choose (in this case, 4), and \( ! \) denotes factorial, the product of all positive integers up to that number. Plugging in the numbers: \[ \binom{30}{4} = \frac{30!}{4!(30-4)!} = \frac{30!}{4! \cdot 26!} \] Now, we can simplify this expression: \[ = \frac{30 \times 29 \times 28 \times 27}{4 \times 3 \times 2 \times 1} \] Calculating the numerator: \[ 30 \times 29 = 870 \] \[ 870 \times 28 = 24360 \] \[ 24360 \times 27 = 657720 \] Now calculating the denominator: \[ 4 \times 3 \times 2 \times 1 = 24 \] Now divide the numerator by the denominator: \[ \frac{657720}{24} = 27405 \] Thus, the number of different ways to select 4 of the 30 numbers is: \[ \boxed{27405} \]