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| 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> |
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| + | '''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> |
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− | ===formulas===
| + | '''Concepts'''<br> |
− | <math>{^{8}}C_{2}</math><br>
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− | <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> |
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− | <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> |
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| =<math>\frac{8!}{6!2!}</math><br> | | =<math>\frac{8!}{6!2!}</math><br> |
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− | =<math>\frac{8X7X6X5X4X3X2X1}{(6X5X4X3X2X1X2X1)}</math><br> | + | =<math>\frac{8X7X6X5X4X3X2X1}{6X5X4X3X2X1X2X1}</math><br> |
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| <math>{^{8}}C_{2}</math>=4X7<br> | | <math>{^{8}}C_{2}</math>=4X7<br> |
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− | <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 |