排列组合问题|套装2

先决条件：排列组合

给定一个m边的多边形，计算可以使用多边形的顶点形成的三角形的数量。

Answer : [m (m – 1)(m – 2) / 6]
Explanation : There are m vertices in a polygon with m sides. We need to count different combinations of three points chosen from m. So answer is ^mC₃ = m * (m – 1) * (m – 2) / 6
Examples :
Input: m = 3
Output: 1
We put value of m = 3, we get required no. of triangles = 3 * 2 * 1 / 6 = 1

Input : m = 6
output : 20

给定一个m边的多边形，计算可以使用多边形的顶点形成的对角线数量。

Answer : [m (m – 3)] / 2.
Explanation : We need to choose two vertices from polygon. We can choose first vertex m ways. We can choose second vertex in m-3 ways (Note that we can not choose adjacent two vertices to form a diagonal). So total number is m * (m – 3). This is twice the total number of combinations as we consider an diagonal edge u-v twice (u-v and v-u)
Examples:
Input m = 4
output: 2
We put the value of m = 4, we get the number of required diagonals = 4 * (4 – 3) / 2 = 2

Input: m = 5
Output: 5

计算可以使用m条垂直线和n条水平线形成的矩形总数

Answer : (^mC₂*ⁿC₂).
We need to choose two vertical lines and two horizontal lines. Since the vertical and horizontal lines are chosen independently, we multiply the result.
Examples:
Input: m = 2, n = 2
Output: 1
We have the total no of rectangles
= ²C₂*²C₂
= 1 * 1 = 1

Input: m = 4, n = 4
Output: 36

平面中有“ n”个点，其中“ m”个点是共线的。找出由这些点形成的三角形的数量为顶点?

Number of triangles = ⁿC₃ – ^mC₃
Explanation : Consider the example n = 10, m = 4. There are 10 points, out of which 4 collinear. A triangle will be formed by any three of these ten points. Thus forming a triangle amounts to selecting any three of the 10 points. Three points can be selected out of the 10 points in ⁿC₃ ways.
Number of triangles formed by 10 points when no 3 of them are co-linear = ¹⁰C₃……(i)
Similarly, the number of triangles formed by 4 points when no 3 of them are co-linear = ⁴C₃……..(ii)

Since triangle formed by these 4 points are not valid, required number of triangles formed = ¹⁰C₃ – ⁴C₃ = 120 – 4 = 116

平面中有“ n”个点，其中“ m”个点是共线的，计算通过连接两个点形成的不同直线的数量。

Answer : ⁿC₂ – ^mC₂ + 1
Explanation : Number of straight lines formed by n points when none of them are col-linear = ⁿC₂
Similarly, the number of straight lines formed by m points when none of them collinear = ^mC₂
m points are collinear and reduce in one line. Therefore we subtract ^mC₂ and add 1.
Hence answer = ⁿC₂ – ^mC₂ +1
Examples:
Input : n = 4, m = 3
output : 1
We apply this formula
Answer = ⁴C₂ – ³C₂ +1
= 3 – 3 + 1
= 1

有关排列和组合的更多实践问题：
排列组合测验
组合与排列练习题