A stick has a length of 5 units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are longer than 1 unit? Answer is NOT 3/5

Here's what seems like the sort of argument you're looking for (based off of a trick Wendel used to compute the probability the convex hull of a set of random points on a sphere contains the center of the sphere, which is really the same question in disguise):

Connect the endpoints of the stick into a circle. We now imagine we're cutting at three points instead of two. We can form a triangle if none of the resulting pieces is at least 1/2, i.e. if no semicircle contains all three of our cut points.

Now imagine our cut as being formed in two stages. In the first stage, we choose three pairs of antipodal points on the circle. In the second, we choose one point from each pair to cut at. The sets of three points lying in a semicircle (the nontriangles) correspond exactly to the sets of three consecutive points out of our six chosen points. This means that 6 out of the possible 8 selections in the second stage lead to a non-triangle, regardless of the pairs of points chosen in the first stage.

Step-by-step explanation:

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

4x and 6y

im not sure though