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?

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: https://brainly.com/question/11732255 #SPJ1 Reply

number of solutionsof a+b+c+d+e+f = 2006 is 9.12 * 10^16## How to determine the number of solutions?

equationis given as:equation, we have:combination formula:number of solutionsof a+b+c+d+e+f = 2006 is 9.12 * 10^16combinationsat: