Explanation

(c)
If 1,2, 3, 4 are the roots of given equation, then
$(x - 1)(x - 2)(x - 3)(x - 4) = x^{4} + ax^{3} + bx^{2} + cx + d$
$\Rightarrow \left( x^{2} - 3x + 2 \right)\left( x^{2} - 7x + 12 \right) = x^{4} + ax^{3} + bx^{2} + cx + d$
$\Rightarrow x^{4} - 10x^{3} + 35x^{2} - 50x + 24$
$= x^{4} + ax^{3} + bx^{2} + cx + d$
$\Rightarrow a = - 10,\ b = 35,\ c = - 50,\ d = 24$
$\therefore\ \ a + 2b + c = - 10 + 2 \times 35 - 50 = 10$
Alternate
Since, 1, 2, 3 and 4 are the roots of the equation
$x^{4} + ax^{3} + bx^{2} + cx + d = 0$, then
$1 + a + b + c + d = 0$ ...(i)
$16 + 8a + 4b + 2c + d = 0$ ...(ii)
$81 + 27a + 9b + 3c + d = 0$ ...(iii)
And $256 + 64a + 16b + 4c + d = 0$ ...(iv)
On solving Eqs. (i), (ii), (iii) and (iv), we get
$a = - 10,\ b = 35,\ c = - 50,\ \ d = 24$
Now, $a + 2b + c = - 10 + 2 \times 35 + ( - 50)$
$= - 10 + 70 - 50 = 10$