My approach:
Since the groups are disjoint, I have to find out the number of ways I can group 8 persons out of 50. Then, multiply it with the number of ways I can take 42 people out of 50 (subtracting the 8 people I already took in a group and making both groups disjoint). Which is = $\binom{50}{8} .\binom{42}{7}$
But the book says: $\dfrac{50!}{8!.7!}$ which is greater than my answer.
Whats wrong with my initial approach?
Your answer seems right to me. And it matches the book's if you put a $35!$ in the denominator. It could simply be a mistake in what is printed.
What if it were two disjoint groups of sizes $1$ and $2$ from $5$ total? Your method gives $\binom{5}{2}\cdot\binom{3}{1}=30$, which is so small it is easy to verify directly.
$$\begin{array}{ccccc} AB,C&AB,D&AB,E&AC,B&AC,D&AC,E\\ AD,B&AD,C&AD,E&AE,B&AE,C&AE,D\\ BC,A&BC,D&BC,E&BD,A&BD,C&BD,E\\ BE,A&BE,C&BE,D&CD,A&CD,B&CD,E\\ CE,A&CE,B&CE,D&DE,A&DE,B&DE,C\\ \end{array}$$
The corresponding book answer would be $\frac{5!}{2!\cdot1!}=60$, and demonstrably wrong.