How many solutions does the equation a+b+c+d+e+f=2006. They are positive integers. Your FINAL answer should be in the form x!/x!•x!, where x is a placeholder

Question

hongcuc2 · Answer

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:  brainly.com/question/11732255    #SPJ1

How to determine the number of solutions?

Leave a Comment Cancel reply