跪求一排列组合题解释～

yongyang86r

32 How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?
a)21
b)170
c)340
d)357
e)420

n(n-3)/2 is the total number of diagonals

Each vertex sends out n-3 diagonals (n being the number of vertices)

so, to answer this question above,

n(n-3)/2 - (x*n-3 ) = (n-3) [n/2 - x]

substitute n with number of sides (which is equal to number of vertices and x as the number of vertices that do not connect)

32 => 18 * [21/2 - 1] = 9*19 = 171 (??? answer choice does not exist??)

Help～～

yongyang86r

以下是引用lanseb8ma在2007-8-2 14:35:00的发言：

use the formula you have provided, if every vertice is connected to the rest of the group, we could get:21*(21-3)/2=189 diagonals. But the question says one of the vertices is not connected to any diagonal. Whichever vertice it is,the situation reduces (21-3)=18 diagonals: 21 sides polygon has 21 vertices, 1 vertice should have 18 diagonals leading to all other vertices except its two adjacent vertices and itself (which add to 3).

Therefore, the answer should be (189-18)=171,would it be possible that the given option is wrong?

That is what i thought! Maybe it is wrong.But the official answer is:

The best answer is B.

We have 20 vertices linking to 17 others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170. The vertex that does not connect to any diagonal is just not counted.

yongyang86r

以下是引用Factiva在2007-8-2 22:15:00的发言：
The right answer should be 171. The "official" explanation illogically turns the question into a polygen with 20 vertices, missing one diagonal.

The "20" in the official answer means the left 20 vertices, not misunderstanding the 21 vertices to 20 vertices.

yongyang86r

So the official answer must be wrong...