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There is only one way to get `x^2,x*x` so you have `1x^2`
There are two ways to get `x^3;x*x^2"and"x^2*x` so `2x^3`
There are 3 ways to get `x^4;x*x^3,x^3*x,x^2*x^2` so `3x^4`
There are 4 ways to get `x^5;x*x^4,x^2*x^3,x^3*x^2,x^4*x^1`
There are 5 ways to get `x^6;x*x^5,x^2*x^4,x^3*x^3,x^4*x^2,x^5*x`
There are 6 ways to get `x^7;x*x^6,x^2*x^5,x^3*x^4,x^4*x^3,x^5*x^2,x^6*x`
There are 5 ways to get `x^8;x^2*x^6,x^3*x^5,x^4*x^4,x^5*x^3,x^6*x^2`
and the pattern continues.
These are the number of ways you can get a sum rolling two dice.
This would be
`x(x+x^2+x^3+x^4+x^5+x^6) + x^2(x+x^2+x^3+x^4+x^5+x^6)+`
4 more terms.
We can see that the first term above has no `x^8` term but the remaining 5 terms have one `x^8` term, this means the coefficient on the `x^8` term would be 5.
`x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + 3x^10 + 2x^11 + x^12` would be the exact answer and can be done without a lot of multiplications.
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