Given 10 points in a plane, 5 of which lie in the same straight line and no other group of 3 or more points can be drawn through by one straight line. Find the number of different triangles that can be formed by using the 10 points as vertices.
Please teach me with explanation...thank you very much!!
maths and stat(combination)
2010-06-15 10:25 pm
回答 (1)
2010-06-15 10:39 pm
✔ 最佳答案
一個平面上有10點﹐其中除5點成一直線﹐其餘任意三點皆不能成一直線。問這10點可以組成多少個不同的三角形?答:
首先因為3點成一三角形﹐這時三角形總數是10C3=120
但由已知直線上5點任取3點的三角形是「直線三角形」要捨去﹐其數目是5C3=10
所以這10點可以組成不同的三角形數目
=10C3-5C3
=120-10
=110
收錄日期: 2021-04-13 17:18:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100615000051KK00689