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Okay given an expression like $(x+y)^5$

we have terms

$x^ky^{n-k}$

So we have ${5 \choose 0} x^0y^5 + {5 \choose 1} x^1y^4+ {5 \choose 2} x^2y^3+ {5 \choose 3} x^3y^2+ {5 \choose 4} x^4y^1+ {5 \choose 5} x^5$

However when we have $(3x^2+y)^5$

we supposedly have $x^{n-k}y^k$ and we get:

${5 \choose 0} (3x^2)^5y^0+ {5 \choose 1} (3x^2)^4y^1+{5 \choose 2} (3x^2)^3y^2+{5 \choose 3} (3x^2)^2y^3 + {5 \choose 4} (3x^2)y^4+ {5 \choose 5} (3x^2)^0y^5$

which ends up as:

$243x^{10}y^0+ 625x^8y+27x^6y^2+9x^4y^3+3x^2y^4+y^5$

somehow my coefficients are off for the second term : it's apparently $5$ times $3^4$. How in the heck does that arise do we always multiply coefficients by the power n, in this case 5 and secondly my biggest question is why on earth is n-k on x instead of y?

What would happen in a situation where we have $(4x^3+5y^2)^3$ which term would get n-k and would I have to multiply some coefficients by 3?

What if we had $(x^2+7y^3)^2$? would n-k go on the $7y^3$?

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    The second term in your expansion of $(3 x^2 + y)^5$ is ${5 \choose 1} (3 x^2)^4 y^1 = \frac{5!}{4!1!} 3^4 (x^2)^4 y = (5 \cdot 3^4) x^8 y$, giving the desired coefficient. – 2017-01-21
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    The term $n-k$ can be taken on any term as $\binom{n}{n-k}=\binom{n}{k}$ and all $x^{n-k}y^k$ are included as $k$ runs from $0,1,...,n$. – 2017-01-21

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Expansion of the expression $\left(x+y\right)^n$ gives :

$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^ky^{n-k}$$

but you can also write it as :

$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$

simply because addition and multiplication are commutative.

So you can put the exponent $n-k$ to whatever term you wish and put the exponent $k$ to the other one.