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§4.对几个问题的演变情况讨论: 1.今举出几个简单例题,说明问题的演变。有时由于条件的不同,要求的不同,使问题变得复杂化,甚至变更问题的性质,因之解决问题的方法,亦就根据它的演变而不同。硬套公式是不好的,有时公式亦无法随便就会套上去。主要在抓紧问题的性质,要求结合习惯加以细致的分析,以决定解决问题的方法。例一、从六个数字0,1,2,3,4,5中每次取出五个来排列: (1)不得取重复的有几种,又可取重复的有几种? 这题很明显的是一个简单排列题。从六个中任取五个的排列数,假设不取重复,有A_6~5=6·5·4·3·2·1=720种。假设可取重复,有6~5=7776种。我们在上面两种做法分析它的不同之处。今可假定有五个位置: (?) (?) (?) (?) (?) 分五步来排满它。在不取重复的条件下,首位有六种方法(在0,1,2,3,4,5)填满它,在首位填过一个之后,第二
§ 4. Discussion of the evolution of several issues: 1. Now give some simple examples to illustrate the evolution of the problem. Sometimes, because of different conditions and different requirements, the problem becomes complicated, and even the nature of the problem is changed. Therefore, the method of solving the problem also differs depending on its evolution. The formula is not good, sometimes the formula can not be set up. Mainly to grasp the nature of the problem, require a detailed analysis combined with the habit to determine the solution to the problem. Example 1, from the six figures 0,1,2,3,4,5 to take out five each time to arrange: (1) There are several kinds of repetitions, and there are several kinds of repetitions that are desirable? This question is obvious Is a simple arrangement of questions. From five out of six, the number of permutations is assumed to be A_6~5=6·5·4·3·2·1=720. Assuming repeatability, there are 6~5=7776 kinds. We analyze its differences in the above two approaches. It can be assumed that there are five positions: (?) (?) (?) (?) (?) to fill it in five steps. In the absence of duplication, there are six methods in the first place (in 0, 1, 2, 3, 4, 5) to fill it, after filling one in the first place, the second