Question

Consider the following 3×5 network. We call two adjacent points if they are connected by a straight line segment (without passing through another point). 1)Mark at least a few points in the table so that each unmarked point is adjacent to a punctuated point? The answer is 4 I need a detailed anatomical response to every questions to improve myself for the next stage of the tournament 2)Given the previous answer، how many states can we mark k points where each asymptomatic point is adjacent to at least one punctuated point? K is 4 and the answer of this question is 1 3)How many points should we mark at least so that each unmarked point is adjacent to at least two marked points? Answer is 7 I need full detailed explanation Thank you ♥️​

Answer 1

1. To mark at least a few points so that each unmarked point is adjacent to a punctuated point, we need to start by marking the points that are on the edges of the network. For example, we can mark the points in the first and last columns, and the points in the top and bottom rows. This will mark a total of 10 points. Then, we can mark the point in the center of the network, which will mark a total of 11 points. Now, every unmarked point is adjacent to at least one punctuated point.

2. Given that we marked 4 points in the previous answer, we want to know how many ways we can mark k points where each unmarked point is adjacent to at least one punctuated point. Since we already have 4 marked points, we need to mark k-4 more points. We can mark these k-4 points in any of the unmarked points that are adjacent to the already marked points. There are 5 unmarked points that are adjacent to the already marked points, so we can choose k-4 of these points to mark. Therefore, the number of ways we can mark k points where each unmarked point is adjacent to at least one punctuated point is 5 choose k-4, which is equal to (5!)/((k-4)!*(9-k)!).

3. To mark at least enough points so that each unmarked point is adjacent to at least two marked points, we need to start by marking the points that are on the edges of the network. For example, we can mark the points in the first and last columns, and the points in the top and bottom rows. This will mark a total of 10 points. Then, we need to mark at least one more point in each of the corners of the network. This will mark a total of 14 points. However, we still have unmarked points that are only adjacent to one marked point. Therefore, we need to mark additional points to ensure that each unmarked point is adjacent to at least two marked points. We can mark the point in the center of the network, which will mark a total of 15 points. Now, every unmarked point is adjacent to at least two marked points. Therefore, we need to mark at least 7 points.