Difference between revisions of "Permutations And Combinations"
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= Hints for difficult problems = | = Hints for difficult problems = | ||
| + | # How many 1) lines` | ||
| + | 2) Triangles can be drawn through 8 points on a circle | ||
| + | |||
| + | Interpetation: | ||
| + | drawing lines and triangle using the 8 points which are on a circumference of a circle (irrespective of the position on circumference , may be equally spaced or not) . And students should identify whether the problem is of arrengement or selection of objects. | ||
| + | |||
| + | Methods: | ||
| + | 1) visual method | ||
| + | 2) arithmetic way | ||
| + | |||
| + | Concepts: | ||
| + | 1. circle | ||
| + | 2. co-linear and non co-linear points | ||
| + | 3. difference bet'n line, ray and segment | ||
| + | 4. points require to draw a triangle | ||
| + | 5. difference bet'n arrengements and selection for eg. If A and B are two objects then AB and BA are different arregements where as AB and BA are both same when it comes to selection. | ||
| + | 6. Expantion of Factorial | ||
| + | |||
| + | Solution: | ||
| + | |||
| + | 1. for drawing lines we need atleast two points then selecting 2 points out of 8 given we have | ||
| + | |||
| + | <br> = arrangements | ||
| + | |||
| + | |||
| + | <br>n! | ||
| + | <br> =----------- ( students should know the formula) | ||
| + | <br>(n-r)! r! | ||
| + | |||
| + | <br> 8! | ||
| + | <br> = ----------- (substituting the values into the formula) | ||
| + | <br> (8-2)! 2! | ||
| + | |||
| + | <br>8! | ||
| + | <br>= ----------- (simplification) | ||
| + | <br>6! 2! | ||
| + | |||
| + | <br> | ||
| + | <br> 8*7*6*5*4*3*2*1 | ||
| + | <br> =-------------------------- | ||
| + | <br> 6*5*4*3*2*1*2*1 | ||
| + | |||
| + | <br> =4*7 | ||
| + | <br> | ||
| + | <br> = 28 | ||
| + | |||
| + | <br> 2. for drawing a triangle we need 3 non co-linear points then selecting 3 points out of 8 given we have | ||
| + | <br> = arrengements | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | <br> n! | ||
| + | <br> =----------- ( students should know the formula) | ||
| + | <br> (n-r)! r! | ||
| + | |||
| + | <br> 8! | ||
| + | <br> = ----------- (substituting the values into the formula) | ||
| + | <br> (8-3)! 3! | ||
| + | |||
| + | <br> 8! | ||
| + | <br>= ----------- (simplification) | ||
| + | <br>5! 3! | ||
| + | |||
| + | <br> | ||
| + | <br> 8*7*6*5*4*3*2*1 | ||
| + | <br> =-------------------------- | ||
| + | <br> 5*4*3*2*1*3*2*1 | ||
| + | |||
| + | <br> =8*7 | ||
| + | |||
| + | <br> = 56 | ||
= Project Ideas = | = Project Ideas = | ||
Revision as of 16:32, 8 July 2014
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permutation and combination [1]
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#Activity No # 1 How to make Permutations "https://www.geogebratube.org/material/iframe/id/11059/width/752/height/329/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5"
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- Activity No #1
- Activity No #2
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Hints for difficult problems
- How many 1) lines`
2) Triangles can be drawn through 8 points on a circle
Interpetation: drawing lines and triangle using the 8 points which are on a circumference of a circle (irrespective of the position on circumference , may be equally spaced or not) . And students should identify whether the problem is of arrengement or selection of objects.
Methods: 1) visual method 2) arithmetic way
Concepts: 1. circle 2. co-linear and non co-linear points 3. difference bet'n line, ray and segment 4. points require to draw a triangle 5. difference bet'n arrengements and selection for eg. If A and B are two objects then AB and BA are different arregements where as AB and BA are both same when it comes to selection. 6. Expantion of Factorial
Solution:
1. for drawing lines we need atleast two points then selecting 2 points out of 8 given we have
= arrangements
n!
=----------- ( students should know the formula)
(n-r)! r!
8!
= ----------- (substituting the values into the formula)
(8-2)! 2!
8!
= ----------- (simplification)
6! 2!
8*7*6*5*4*3*2*1
=--------------------------
6*5*4*3*2*1*2*1
=4*7
= 28
2. for drawing a triangle we need 3 non co-linear points then selecting 3 points out of 8 given we have
= arrengements
n!
=----------- ( students should know the formula)
(n-r)! r!
8!
= ----------- (substituting the values into the formula)
(8-3)! 3!
8!
= ----------- (simplification)
5! 3!
8*7*6*5*4*3*2*1
=--------------------------
5*4*3*2*1*3*2*1
=8*7
= 56
Project Ideas
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