If 1, 2, 3 and 4 are the roots of the equation ${\mathrm{x}}^4 + {\mathrm{ax}}^3 + {\mathrm{bx}}^2 + \mathrm{cx} + \mathrm{d} = 0 \mathrm{then} \mathrm{a} + 2\mathrm{b} + \mathrm{c} =$

$\begin{array}{l}TheEquationhavig1,2,3,4arerootsis\\ (x-1)(x-2)(x-3)(x-4)=0\\ \Rightarrow (x^2-3x+2)(x^2-7x+12)=0\\ \Rightarrow x^4-7x^3+12x^2-3x^3+21x^2-36x+2x^2-14x+24=0\\ \Rightarrow x^4-10x^3+35x^2-50x+24=0\\ Comparingwith{x}^4+a{x}^3+b{x}^2+cx+d=0\\ a=-10,b=35,c=-50\\ \therefore a+2b+c=-10+70-50\Rightarrow 10\end{array}$