Out of 10 contestants, six are to be selected for the final round of a competition - HSC - SSCE Mathematics Extension 1 - Question 10 - 2020 - Paper 1

Next, we need to arrange 4 of the selected contestants into 1st, 2nd, 3rd, and 4th places. The number of arrangements (permutations) is given by:

$P(n, r) = \frac{n!}{(n-r)!}$

For our purpose, we have:

$P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!}$

The total number of ways to select and arrange the contestants is the product of the two calculated steps:

$\text{Total Ways} = C(10, 6) \cdot P(6, 4) = \frac{10!}{6!4!} \cdot \frac{6!}{2!} = \frac{10!}{4! imes 2!}$

Thus, the final answer provides options where:

A: $\frac{10!}{6!4!}$

Since this simplifies accordingly, the answer is: Option A.