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 11:02, 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"

  1. Activity No #2

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Activities

  1. Activity No #1
  2. Activity No #2

Assessment activities for CCE

Hints for difficult problems

  1. 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

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