To determine the number of ways an elevator can stop at three distinct floors out of 10, we can use the combinations formula. In mathematics, combinations refer to the selection of items from a larger set, such that the order of the items does not matter.
The formula for combinations is:
$$C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$
where C(n, r), or $\binom{n}{r}$, is the number of combinations, n is the total number of items in the set, r is the number of items we want to choose from the set, and n!, r!, and (n − r)! represent the factorial of n, r, and n − r, respectively. The factorial of a number x, denoted by x!, is the product of all positive integers less than or equal to x.
In our problem, there are a total of 10 floors (i.e., n = 10), and we want to choose 3 floors for the elevator to stop (i.e., r = 3).
Applying the combinations formula:
$$C(10, 3) = \binom{10}{3} = \frac{10!}{3!(10-3)!}$$
Now, we need to evaluate the factorials and the expression:
$$\frac{10!}{3!(7)!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)((7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1))}$$
We can now simplify this expression:
$$\frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{10 \times 3 \times 3 \times 8}{3 \times 2 \times 1} = 120$$
So, there are 120 ways for the elevator to stop at any three distinct floors out of the 10 floors.