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  • find the number of 3 digit numbers,whose middle digit is bigger than extreme digits
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  • well.....that was a good question indeed...

    answer is 240


    least value of middle number: 2

    Units digit can have 2 possible values (0,1)
    and the 100th digit can have 1 possible value (1) therefore, numbers possible is 1X2

    When, middle number's value is 3. Units digit can have 3 possible values (0,1,2) and the 100th digit can have 2 possible values (1,2) therefore, numbers possible is 2X3

    And so on....(units digit value could be 2....9)

    Hence, total no. of possible numbers are: 1X2 + 2X3 + .........+ 7X8 + 8X9 = 240


    Karma me +1 if im correct!! :-)
  • the no of triangles that can be formedby 5 points in a line and 3 points on a parallel line
  • 8C3-5C3-3C3
  • i think viniket is rite!!

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