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2 years ago

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?

Mathematics

1 answer:

Sati [7]2 years ago

3 0

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

<h3>How to determine the number of solutions?</h3>

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

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