Question

How many solutions does the equation a+b+c+d+e+f = 2006 have where a b c d e and f are all positive integers?

Answers

  1. The number of solutions of a+b+c+d+e+f = 2006 is 9.12 * 10^16

    How to determine the number of solutions?

    The equation is given as:
    a+b+c+d+e+f = 2006
    In the above equation, we have:
    Result = 2006
    Variables = 6
    This means that
    n = 2006
    r = 6
    The number of solutions is then calculated as:
    (n + r – 1)Cr
    This gives
    (2006 + 6 – 1)C6
    Evaluate the sum and difference
    2011C6
    Apply the combination formula:
    2011C6 = 2011!/((2011-6)! * 6!)
    Evaluate the difference
    2011C6 = 2011!/(2005! * 6!)
    Expand the expression
    2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006 * 2005!/(2005! * 6!)
    Cancel out the common factors
    2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006/6!
    Expand the denominator
    2011C6 = 2011 * 2010 * 2009 * 2008 * 2007 * 2006/720
    Evaluate the quotient
    2011C6 = 9.12 * 10^16
    Hence, the number of solutions of a+b+c+d+e+f = 2006 is 9.12 * 10^16
    Read more about combinations at:
    #SPJ1

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