1) 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.
2) A transportation company has 8 lorries and 10 drivers available. When it receives 4 orders of one lorry each, how many different selections of lorries and drivers are there to meet the requests?
Combination
2009-09-12 10:55 pm
回答 (2)
2009-09-12 11:30 pm
✔ 最佳答案
(1) Consider no three points are collinear, there are 10C3 triangles possible.When 3 points are collinear, we cannot form a triangle. Since there are 5 points that are collinear, 5C3 should be removed from the calculation.
Number of triangles possible = 10C3 - 5C3 = 120 - 10 = 110
(2) There are 8C4 ways to select the lorries.
There are 10C4 ways to select the drivers.
Total possible ways is (8C4)(10C4) = (70)(210) = 14,700 lorry-driver combinations.
To match each of these lorry-driver combination with each of the orders, there are 4!*14700 = 352800 ways
2009-09-12 11:46 pm
1)
case 1.
2 points are chosen from the 5 points on the line.
1 point is chosen from other 5 points.
no. of triangles
= (5C2)x5
= [(5x4)/(2x1)]x5
= 50
case 2.
only 1 point is chosen from the 5 points on the line.
no. of triangles
= 5x(5C2)
= 50
case 3.
all 3 points are chosen from the 5 points not on the line.
no. of triangles
= (5C3)
= (5x4x3)/(3x2x1)
= 10
so, total no. of triangles can be drawn
= 50+50+10
= 110
2)
No. of ways to choose 4 lorries
= (8C4)
= (8x7x6x5)/(4x3x2x1)
= 70
No. of ways to choose 4 drivers
= (8C4)
= 70
No. of ways for drivers to choose their own car
= 4x3x2x1
= 24
so no. of selection ways
= 70x70x24
= 117600
case 1.
2 points are chosen from the 5 points on the line.
1 point is chosen from other 5 points.
no. of triangles
= (5C2)x5
= [(5x4)/(2x1)]x5
= 50
case 2.
only 1 point is chosen from the 5 points on the line.
no. of triangles
= 5x(5C2)
= 50
case 3.
all 3 points are chosen from the 5 points not on the line.
no. of triangles
= (5C3)
= (5x4x3)/(3x2x1)
= 10
so, total no. of triangles can be drawn
= 50+50+10
= 110
2)
No. of ways to choose 4 lorries
= (8C4)
= (8x7x6x5)/(4x3x2x1)
= 70
No. of ways to choose 4 drivers
= (8C4)
= 70
No. of ways for drivers to choose their own car
= 4x3x2x1
= 24
so no. of selection ways
= 70x70x24
= 117600
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