We need to calculate the coefficient of $x^5$ in the expression:

$(1 + x^3)^{15} \cdot (1 + x^2)^{10}.$

Step-by-step Solution:

Step 1: Expand both terms

Using the binomial theorem, each term expands as follows:

1. $(1 + x^3)^{15} = \sum_{k=0}^{15} \binom{15}{k} x^{3k}$,
2. $(1 + x^2)^{10} = \sum_{m=0}^{10} \binom{10}{m} x^{2m}$.

The product of these expansions is:

$(1 + x^3)^{15}(1 + x^2)^{10} = \sum_{k=0}^{15} \sum_{m=0}^{10} \binom{15}{k} \binom{10}{m} x^{3k + 2m}.$

Step 2: Condition for $x^5$

We are interested in the coefficient of $x^5$. This occurs when:

$3k + 2m = 5.$

Step 3: Solve the equation $3k + 2m = 5$

We now find all integer pairs $(k, m)$ that satisfy this equation:

1. For $k = 1$: $3(1) + 2m = 5 \implies 2m = 2 \implies m = 1$.
2. For $k = 0$: $3(0) + 2m = 5 \implies 2m = 5 \implies m$ is not an integer.
3. For $k = 2$: $3(2) + 2m = 5 \implies m = -\frac{1}{2}$, which is not valid.

Thus, the only valid solution is $(k, m) = (1, 1)$.

Step 4: Coefficient of $x^5$

For $(k, m) = (1, 1)$, the coefficient is:

$\binom{15}{1} \cdot \binom{10}{1} = 15 \cdot 10 = 150.$

Final Answer:

The coefficient of $x^5$ is $\mathbf{150}$.

Let me know if you want detailed explanations of any step. Here are related questions to explore:

1. What is the general form of the expansion of a binomial?
2. How does the binomial theorem apply to negative exponents?
3. Can you calculate coefficients for other powers in this expression?
4. How can multinomial expansion help in solving similar problems?
5. What is the significance of Pascal's triangle in such calculations?

Tip: Always double-check if all constraints (e.g., integer solutions) are satisfied when solving equations for coefficients.