本帖最后由 崔老师 于 2023-5-20 09:49 编辑
There are 30 questions in the NMOS exam, with the majority ofthem testing proficiency, while a few difficult questions require strongproblem-solving skills. The following is the distribution of challengingquestions in NMOS exams from 2006 to 2021. From the graph, we can see that counting problems have thehighest frequency among the difficult questions. If you want to score well, youneed to thoroughly understand counting problems! Today, I will talk about themost difficult types of counting problems in NMOS exams. Ⅰ. Permutation CountingProblems Let's start with a pastquestion:
(NMOS 2017R1 Q30) A 4-digit number is formed by using theintegers only once in the list {1, 3, 4, 6, 8, 9} For example, 1348, 3691, and4896 are possible numbers, but 1234 and 3344 are not. If the numbers arearranged from the smallest to the biggest, which position would the number 4386be in? In this question, we are given a number and asked to determineits position. Here's the detailed explanation: For four-digit numbers smallerthan 4386, the first digit must be 1, 3, or 4. When the first digit is 1 or 3,the other digits can be filled arbitrarily. There are 2 options for the firstdigit, and the other three digits have 5, 4, and 3 options respectively. So,there are a total of 2 × 5 × 4 × 3 = 120 numbers. When the first digit is 4,the hundreds digit can only be 1 or 3. If the hundreds digit is 1, the tens andunits digits have 4 and 3 options respectively, resulting in 4 × 3 = 12numbers. If the hundreds digit is 3, the tens and units digits can be 16, 18,19, 61, 68, 69, or 81, resulting in a total of 7 numbers. Therefore, the totalnumber of four-digit numbers smaller than 4386 is 120 + 12 + 7 = 139. Thus, thenumber 4386 is in the 140th position.
The key to this method is to classify the numbers inthe list based on the permutation pattern and calculate each categoryseparately!
Using this approach, let's quickly solve another question: (NMOS 2009R1 Q29) Kevin wrote whole numbers starting from 1.Then he put a "#" behind every 3 digits as follows: 1 2 3 # 4 5 6 # 78 9 # 1 0 1 # 1 1 2 # 1 3 1 # ... What is the digit before the 100th"#"?
In contrast to the previous question, this time we are given theposition and asked to find the digit. However, the approach is similar:Firstly, the 100th "#" corresponds to the 300th digit in the sequence.We can classify the digits based on the number of positions each multi-digitnumber occupies: Single-digit numbers: 1 to 9, a total of 9 positions.Two-digit numbers: 10 to 99, a total of 2 × 90 = 180 positions. Three-digitnumbers: The remaining 300 - 180 - 9 = 111 positions, which is equivalent to111 ÷ 3 = 37 three-digit numbers. Therefore, we have exactly 99 + 37 = 136numbers written so far. Thus, the digit before the 100th "#" is 6.
For these permutation counting problems, the key is to classify thenumbers in the sequence based on the permutation patterns. Withproper classification, solving them becomes easy. Ⅱ. Approach from the OppositePerspective Let's start with a past question: (NMOS 2009R1 Q29) How many whole numbers from 1 to 2009 containthe digit "2"?
The question is straightforward, but if we carefully considerthe numbers from 1 to 2009 that contain the digit "2," we'll realizethat the situation is quite complex. How many occurrences of the digit"2" are there? In which positions does it appear? There are manyclassification scenarios, making the calculation challenging. Therefore,instead of focusing on the numbers that contain "2," we can approachit from the opposite perspective—calculate the numbers that do not contain"2" and subtract them from the total possibilities. The numbers thatdo not contain "2" are equivalent to the range from 1 to 1999. Here'sa technique called "digit extension": we add leading zeros to numberswith less than 4 digits to make them four digits long. For example, 1 can bewritten as 0001, and 45 can be written as 0045. In this way, the count ofnumbers from 1 to 1999 that do not contain "2" is obtained. There are2 options for the first digit from left to right, and the next three digits canbe filled with any number except 2, resulting in a total of 2 × 9 × 9 × 9 =1458 numbers. However, this count includes 0000, so we need to subtract 1,resulting in a total of 1457 numbers that do not contain "2." Therefore, the count of numbers that contain the digit "2" is 2009 -1457 = 552. The key to this question is to approach it from theopposite perspective, considering the complexity of the direct approach. Digitextension can be used to ensure consistent digit lengths.
Ⅲ. Counting Triangles Problem NMOS exams often include questions about counting shapes, withcounting triangles being the most challenging. Here, I'll introduce a"universal method" for counting triangles. Let's look at a pastquestion: (NMOS 2018R1 Q16) Find the total number of triangles in the figureshown below. One approach for this question is to enumerate and classify thetriangles based on the number of regions forming them. However, it is difficultto ensure accuracy without missing or duplicating triangles, especially withmore complex figures. We can simplify the figure as follows: ① In the figure, don't forget the large triangle, which accountsfor 5 triangles. Next, let's add one line: ② In this figure, the additional line introduces new triangles.Each new triangle is formed by the new line and one additional point. From each point in the figure, there are only 2 other pointsthat can form triangles with the red line, resulting in 1 triangle for eachpoint: So far, we have a total of 5 + 2 = 7 triangles. Now, let's add the final line: ③ In this figure, we encounter the following situation: In this case, there are 3 triangles that can be formed by theupper vertex and the red line. We mark each new triangle corresponding to each point: Therefore, the total number of triangles in the originalquestion is 7 + 1 + 3 + 1 + 3 = 15. The key to this question is to simplifythe figure to an easily countable scenario, then systematically add one line ata time, accumulating the newly formed triangles. This method iscalled the "line addition" method. Once you become proficient withthe line addition method, solving challenging triangle counting problemsbecomes much easier!
For students with good foundations, NMOS exams are certainlyaimed at achieving full marks. Therefore, it's crucial to be proficient insolving simple questions and to know and apply problem-solving methods fordifficult questions. This post introduced the three most challenging types ofcounting problems in NMOS exams, and I hope you found it helpful.
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