A certain lottery has 31 numbers. In how many different ways can 6 of the numbers be selected? (Assume that order of selection is not important.) There are different ways the numbers can be selected. (Simplify your answer.)

Step 1 ：This problem is about combinations. In a lottery, the order of the numbers does not matter, so we are looking for the number of combinations of 6 numbers from a pool of 31.

Step 2 ：The formula for combinations is: \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to choose, and '!' denotes factorial, which is the product of all positive integers up to that number.

Step 3 ：In this case, n = 31 (the total number of lottery numbers) and k = 6 (the number of numbers to choose).

Step 4 ：Substituting the values into the formula, we get \(C(31, 6) = \frac{31!}{6!(31-6)!}\).

Step 5 ：Solving this gives us the number of combinations as 736281.

Step 6 ：Final Answer: The number of different ways 6 numbers can be selected from a pool of 31 is \(\boxed{736281}\).