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From Karnataka Open Educational Resources
Activity-permutations and combinations problems (view source)
Revision as of 06:08, 11 July 2014
, 06:08, 11 July 2014no edit summary
Hence by fundamental principle of counting total number of ways =1X5X4X3=60 numbers
Hence by fundamental principle of counting total number of ways =1X5X4X3=60 numbers
=Problem 3=
'''How many 1) lines 2) Triangles can be drawn through 8 points on a circle'''<br>
'''Interpretation''' <br>
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) .students should identify whether the problem is of arrengement or selection of objects.<br>
'''Concepts'''<br>
<math>{^{n}}C_{r}</math>=<math>\frac{n!}{(n-r)!r!}</math><br>
#Circle<br>
#Co-linear and non co-linear points<br>
#Difference bet'n line, ray and segment<br>
#Points require to draw a triangle<br>
#Difference bet'n arrangements and selection for eg. If A and B are two objects then AB and BA are different arrangements where as AB and BA are both same when it comes to selection.<br>
#Expantion of Factorial notation <br>
<math>{^{8}}C_{2}</math>=<math>\frac{8!}{(8-2)!2!}</math><br>
'''Solution'''<br>
*For drawing lines we need atleast two points then selecting 2 points out of 8 given we have
<math>{^{8}}C_{2}</math> combinations<br>
We have <math>{^{n}}C_{r}</math>=<math>\frac{n!}{(n-r)!r!}</math><br>
therefore <math>{^{8}}C_{2}</math>=<math>\frac{8!}{(8-2)!2!}</math><br>
=<math>\frac{8!}{6!2!}</math><br>
=<math>\frac{8!}{6!2!}</math><br>
=<math>\frac{8X7X6X5X4X3X2X1}{(6X5X4X3X2X1X2X1)}</math><br>
=<math>\frac{8X7X6X5X4X3X2X1}{6X5X4X3X2X1X2X1}</math><br>
<math>{^{8}}C_{2}</math>=4X7<br>
<math>{^{8}}C_{2}</math>=4X7<br>
<math>{^{8}}C_{2}</math>=28
<math>{^{8}}C_{2}</math>=28 numbers
*For drawing a triangle we need 3 non co-linear points hence selecting 3 points out of 8 given we have<br>
<math>{^{8}}C_{3}</math> combinations<br>
We have <math>{^{n}}C_{r}</math>=<math>\frac{n!}{(n-r)!r!}</math><br>
therefore <math>{^{8}}C_{3}</math>=<math>\frac{8!}{(8-3)!3!}</math><br>
=<math>\frac{8!}{5!3!}</math><br>
=<math>\frac{8X7X6X5X4X3X2X1}{5X4X3X2X1X3X2X1}</math><br>
<math>{^{8}}C_{2}</math>=8X7<br>
<math>{^{8}}C_{2}</math>=56 numbers
