The coefficient of $$x^{10}$$ in the expansion of $$(1+x)^2(1+x^2)^3(1+x^3)^4$$ is equal to:

We have to find the coefficient of $$x^{10}$$ in the product

$$\left(1+x\right)^2\left(1+x^2\right)^3\left(1+x^3\right)^4.$$

First, we expand each factor completely because that will let us see every possible power along with its coefficient.

For the first factor we use the Binomial Theorem, $$\left(1+x\right)^2=\sum_{r=0}^{2}\binom{2}{r}x^{\,r},$$ so

$$\left(1+x\right)^2 = 1 + 2x + x^2.$$

For the second factor, again using the Binomial Theorem but remembering that the term in the bracket is $$x^2,$$ we get

$$\left(1+x^2\right)^3=\sum_{s=0}^{3}\binom{3}{s}x^{\,2s} = 1 + 3x^2 + 3x^4 + x^6.$$

For the third factor we do the same, treating the basic term as $$x^3:$$

$$\left(1+x^3\right)^4=\sum_{t=0}^{4}\binom{4}{t}x^{\,3t} = 1 + 4x^3 + 6x^6 + 4x^9 + x^{12}.$$

Now we must pick one term from each of the three expanded brackets such that the total power of $$x$$ is $$10.$$ Let the powers chosen from the three factors be:

$$e_1\;(from\;(1+x)^2),\quad e_2\;(from\;(1+x^2)^3),\quad e_3\;(from\;(1+x^3)^4).$$

Because the powers in the second factor come in multiples of 2 and in the third factor in multiples of 3, the only values those exponents can assume (visible in the expansions above) are

$$e_1\in\{0,1,2\},\quad e_2\in\{0,2,4,6\},\quad e_3\in\{0,3,6,9,12\}.$$

We require

$$e_1 + e_2 + e_3 = 10.$$

We check each possible choice of $$e_3$$ one by one and see whether suitable $$e_1,e_2$$ exist.

1. If $$e_3 = 0,$$ then $$e_1+e_2=10,$$ but the largest $$e_2$$ available is $$6,$$ so this case is impossible.

2. If $$e_3 = 3,$$ then $$e_1+e_2=7.$$ Because $$e_2$$ must be even, the only even value near 7 is $$6;$$ that forces $$e_1=1.$$ So one valid triple is $$(e_1,e_2,e_3)=(1,6,3).$$

3. If $$e_3 = 6,$$ then $$e_1+e_2=4.$$ The even options for $$e_2$$ are $$0,2,4.$$   • Taking $$e_2=4$$ gives $$e_1=0.$$br />   • Taking $$e_2=2$$ gives $$e_1=2.$$br /> $$e_2=0$$ would force $$e_1=4,$$ which is not allowed, so we get two more triples: $$(e_1,e_2,e_3)=(0,4,6)\quad\text{and}\quad(2,2,6).$$

4. If $$e_3 = 9,$$ then $$e_1+e_2=1.$$ Since $$e_2$$ is even, the only choice is $$e_2=0,$$ giving $$e_1=1.$$ Thus we obtain $$(e_1,e_2,e_3)=(1,0,9).$$

5. If $$e_3 = 12,$$ then $$e_1+e_2=-2,$$ impossible.

Hence the only admissible exponent triples are

$$ \begin{aligned} (1,6,3),\quad (0,4,6),\quad (2,2,6),\quad (1,0,9). \end{aligned} $$

For each triple we multiply the corresponding coefficients taken from the three expansions.

For $$(e_1,e_2,e_3)=(1,6,3):$$ coefficient $$= \color{blue}{2}\times\color{blue}{1}\times\color{blue}{4}=8.$$ (The numbers in blue are from $$2x,\;x^6,\;4x^3.$)

For $$(e_1,e_2,e_3)=(0,4,6):$$ coefficient $$= 1$$\times$$3$$\times$$6 = 18.$$

For $$(e_1,e_2,e_3)=(2,2,6):$$ coefficient $$= 1$$\times$$3$$\times$$6 = 18.$$

For $$(e_1,e_2,e_3)=(1,0,9):$$ coefficient $$= 2$$\times$$1$$\times$$4 = 8.$$

Finally, we add all these contributions because each gives an independent term with power $$10$$:

$$8 + 18 + 18 + 8 = 52.$$

Hence, the coefficient of $$x^{10}$$ in the given expansion is $$52.$$ Therefore, the correct answer is Option A.