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.

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?

Re: Number of Solutions help

Now the number of solutions is the bonomial coefficient C(4+4-1,4-1)=C(7,3)=35

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 $\displaystyle x^{12}$ in this expansion.

Re: Number of Solutions help