Step 1 :This problem is about calculating the number of combinations. In combinatorics, a combination is a selection of items without considering the order. In this case, we are selecting 6 numbers out of 55, and the order does not matter.
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 = 55 (the numbers from 1 to 55) and k = 6 (we need to choose 6 numbers). So we need to calculate C(55, 6).
Step 4 :By substituting the values into the formula, we get: \(C(55, 6) = \frac{55!}{6!(55-6)!}\)
Step 5 :After calculating the above expression, we find that the number of different selections possible is 28989675.
Step 6 :Final Answer: The number of different selections possible is \(\boxed{28989675}\)