# Number of Solutions help

• Nov 15th 2012, 10:17 PM
gfbrd
Number of Solutions help
Hey I need some help with this problem:

How many solutions in natural numbers are there to the equation a + b + c + d = 12 where a and b are odd?

I am hoping someone can show me the steps on how to go about doing this so I will understand how to do this problem thanks.
• Nov 16th 2012, 04:23 AM
hedi
Re: Number of Solutions help
substitute a=2A+1,b=2B+1,c=2C+1 d=2D+1 and consider the number of non negative integer solutions of A+B+C+D=4.
• Nov 16th 2012, 04:28 PM
gfbrd
Re: Number of Solutions help
Ok so I substituted and ended up with
A+B+C+D = 4 so the possible solutions for a b c d is 1?
• Nov 17th 2012, 04:52 AM
hedi
Re: Number of Solutions help
Now the number of solutions is the bonomial coefficient C(4+4-1,4-1)=C(7,3)=35
• Nov 17th 2012, 05:25 AM
Plato
Re: Number of Solutions help
Quote:

Originally Posted by hedi
substitute a=2A+1,b=2B+1,c=2C+1 d=2D+1 and consider the number of non negative integer solutions of A+B+C+D=4.

You should note that the OP requires only a & b be odd.
Look at the coefficient of $x^{12}$ in this expansion.
• Nov 17th 2012, 07:34 PM
gfbrd
Re: Number of Solutions help
Thanks Plato