Hmm, I've have never seen something like that before

The given expression is $(x-1)(x^2-2)....(x^9-9)(x^{10}-10)$ has degree $\dfrac{(10)\times (11)}{2}=55$ ,but we want $x^{48}$ ,so instead of taking all the terms we exclude those terms,which have the sum of powers of $x=7$ ,because $55-7=48$ ,the solutions are, $\text{Removing}\ x^7 \text{term}\\ \text{Removing}\ x^1,x^6\ \text{terms}\\ \text{Removing}\ x^2,x^5\ \text{terms}\\ \text{Removing}\ x^3,x^4\ \text{terms}\\ \text{Removing}\ x^1,x^2,x^4\ \text{terms}$ .Now let us look at the first case, $(x^7-7)[\text{rest}]$ ,the coefficient of $x^{48}$ will be $-7$ ,similarly in the next cases,the coefficients would be, $(-1)\times(-6)=6,(-2)(-5)=10,(-3)(-4)=12,(-1)(-2)(-4)=8$ ,finally the answer is $-7+6+10+12-8=13$ ,now, $13=2\times 6+1$ ,hence $n=6$ .And done!